Iff Statement Negation Watch

Chittesh14
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Hey,

I was just wondering if someone could help me out here please.
I've got an iff statement in the form p ↔ q.

So, I want to prove the iff statement is true by contradiction, so I am assuming it is false i.e. ¬ (p ↔ q) = ¬ ((p → q) ∧ (q → p)).
= ((p ∧ (¬q)) ∨ (q ∧ (¬p)).
So, that is the negation of (p ↔ q) .
Now, usually - we reach a contradiction e.g. if the form is p implies q, we negate it and get p and (not q) and then we reach a contradiction, showing p and not q is false.
So, for me to have the negation: ((p ∧ (¬q)) ∨ (q ∧ (¬p)).
I need to reach a contradiction here to show that the original statement i.e. (p ↔ q) is true.

To reach a contradiction, do I just have to show that either one of (p ∧ (¬q) or q ∧ (¬p) is false? Would that be a contradiction?
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ghostwalker
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(Original post by Chittesh14)
Hey,

I was just wondering if someone could help me out here please.
I've got an iff statement in the form p ↔ q.

So, I want to prove the iff statement is true by contradiction, so I am assuming it is false i.e. ¬ (p ↔ q) = ¬ ((p → q) ∧ (q → p)).
= ((p ∧ (¬q)) ∨ (q ∧ (¬p)).
So, that is the negation of (p ↔ q) .
Now, usually - we reach a contradiction e.g. if the form is p implies q, we negate it and get p and (not q) and then we reach a contradiction, showing p and not q is false.
So, for me to have the negation: ((p ∧ (¬q)) ∨ (q ∧ (¬p)).
I need to reach a contradiction here to show that the original statement i.e. (p ↔ q) is true.

To reach a contradiction, do I just have to show that either one of (p ∧ (¬q) or q ∧ (¬p) is false? Would that be a contradiction?
You need to show that both parts lead to a contradiction.

I.e. (p ∧ (¬q)) leads to a contraction, and
(q ∧ (¬p)) leads to a contradiction.
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Chittesh14
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(Original post by ghostwalker)
You need to show that both parts lead to a contradiction.

I.e. (p ∧ (¬q)) leads to a contraction, and
(q ∧ (¬p)) leads to a contradiction.
Oh wow, that's beautiful. just wondering why both parts?
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ghostwalker
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(Original post by Chittesh14)
Oh wow, that's beautiful. just wondering why both parts?
If the first part leads to a contradiction, then it must be false, BUT the second part could still be true.

Only by showing that both parts, call them A and B, are false, can you conclude that "A or B" is false.
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Chittesh14
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(Original post by ghostwalker)
If the first part leads to a contradiction, then it must be false, BUT the second part could still be true.

Only by showing that both parts, call them A and B, are false, can you conclude that "A or B" is false.
Oh yes, thank you so much. That is regarding the truth table for 'OR', didn't recognise it - thank you.

In regards to the question I had (which I didn't post), I understand it is not linked to the thread, so if you can answer it that'd be great and much appreciated (because you usually know the answer to all my questions lol), else I'll create a separate thread on it.

So the question was:

For any non-empty bounded set S, prove that inf S ≤ sup S, and that the equality holds iff S is a singleton set (that is a set with exactly one element).

So, I may have interpreted this question incorrectly and if I have please let me know .

So I interpreted this as an if and only if statement where both of these conditions have to be satisfied:

If inf S ≤ sup S, then S is a singleton set
and If S in a singleton set, inf S ≤ sup S.

So I could either prove both of these statements true, or use proof by contradiction in which I would assume the negation of their conjunction and reach a contradiction, right (I think lolz)?

So, proof by contradiction:

Steps of negation
Spoiler:
Show
negation of : {If inf S ≤ sup S, then S is a singleton set and If S in a singleton set, inf S ≤ sup S}
= {negation of (If inf S ≤ sup S, then S is a singleton set) or negation of (If S in a singleton set, inf S ≤ sup S)}
= {inf S ≤ sup S and S is not a singleton set or S in a singleton set and inf S ≥ sup S}


Now, to reach a contradiction: I have to prove both of the statements wrong:

inf S ≤ sup S and S is not a singleton set 1
and
S in a singleton set and inf S > sup S. 2

I mean, I don't really know how this helps lol! But, I just used it as my method.

It's basically the same approach as I would use with (proving both the original statements to be true).

PROOF:

1

In a set, inf(S) and sup(S) are either included in the set or they're not.
In the cases, either the set tends towards a lowest/greatest element, or has a lowest/greatest element.
Let that smallest element that the set tends towards or includes, be a, and let the largest element that the set tends towards or includes, be b.

So lower bound ≤ a (regardless of whether it is included or not), so inf(S) = a. Similarly, upper bound ≥ b (regardless of whether it is included or not), so sup(S) = b.

Now, if there is more than 1 element in the set, the elements have to be distinct, so a ≠ b. As a result, inf (S) ≠ sup (S). As it is not equal to, the equality doesn't hold and therefore inf (S) < sup (S) as it can never be greater than it. We have reached a contradiction, and this statement is false.

2

Using the information in the above (since I don't want to make it really long), if the set only contains one element, then the smallest and the largest element will either be the same, or it will have a smallest element and tends towards a largest element, or it will have a largest element and tends towards a smallest element unsure about these last two.

Is this wrong? E.g. of it will have a smallest element and tends towards a largest element: Let x = {X is an integer | 1 ≤ x < 2}. It doesn't tend towards 2 right because x is not a real number, it is an integer? So would this be incorrect?

If it is incorrect, I'll just prove the first statement: a set where the smallest and the largest element will be the same e.g. a.
In this case, any element ≤ a or ≥ a will be a lower bound and upper bound respectively. So, inf (S) = sup (S) = a.
In this case, the inequality holds and therefore, the strict inequality isn't satisfied - we have reached a contradiction and therefore this statement is false.

As both statements are false, this leads us to conclude that the original statement is true and therefore inf S ≤ sup S, and that the equality holds iff S is a singleton set (that is a set with exactly one element).

By the way, I know you'll be really tired by the time you read all of that, so if you have time, I'd really appreciate if you can just confirm the next part:

Spoiler:
Show


Contrapositive:


I assume what I've written below is the contrapositive, can you please confirm that?

For any non-empty bounded set S, prove that S is not a singleton set (that is a set with exactly one element) iff inf S > sup S, and that the equality doesn't hold, only in this case.

Would this be easier to prove in any way, or is it just the same approach but sort of reversed (like backwards) i.e. proving that if inf S > sup S, then S is not a singleton set ... etc.


Thank you so much for your help as always .
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ghostwalker
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(Original post by Chittesh14)
...
Haven't got the energy at present to even fully read through that, so a general statement:

Iff statements are usually proved by proving the logically equaivalant "(A impiles B) AND (B implies A)" and each part is treated separately.

It is often the case that doing the implication one way is straight forward, almost trivial. And doing the implication the other way is somewhat harder and this part is often best done via proof by contradiction.

All that, rather than negating the original iff in using proof by contradtion on the whole thing.
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DFranklin
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I agree with Ghostwalkers advice, but I think some comments on what you've done are necessary. I've commented inline, in red. It should be said that a lot of what I haven't commented on isn't really "right" either, but there's so many errors that often your subsequent work just doesn't make sense.

(Original post by Chittesh14)
So the question was:

For any non-empty bounded set S, prove that inf S ≤ sup S, and that the equality holds iff S is a singleton set (that is a set with exactly one element).

So, I may have interpreted this question incorrectly and if I have please let me know .

So I interpreted this as an if and only if statement where both of these conditions have to be satisfied:

If inf S ≤ sup S, then S is a singleton set
and If S in a singleton set, inf S ≤ sup S.
This isn't actually correct. You actually have three statements you need to prove:

If S is a non-empty bounded set, inf S ≤ sup S.
If inf S = sup S then S is a singleton set.
If S is a singleton set then inf S = sup S.

The "iff" part of the question only concerns the case where equality holds (i.e. inf S = sup S).

This means most of your subsequent work is invalid, but I have some comments anyhow...
.

~snip~

In a set, inf(S) and sup(S) are either included in the set or they're not.
In the cases, either the set tends towards a lowest/greatest element, or has a lowest/greatest element.
Let that smallest element that the set tends towards or includes, be a, and let the largest element that the set tends towards or includes, be b. What's all this "set tends towards" nonsense? There's a very informal way in which it might make sense, but I can't believe it's any part of your formal definition of inf(S) and sup(S) and it's certainly too informal to be applicable here. .

So lower bound ≤ a (regardless of whether it is included or not), so inf(S) = a. Similarly, upper bound ≥ b (regardless of whether it is included or not), so sup(S) = b.

Now, if there is more than 1 element in the set, the elements have to be distinct, so a ≠ b. If a and b are just things the set "tends towards", I don't see how you can claim this. (Or at least, you definitely need to justify it). .As a result, inf (S) ≠ sup (S). As it is not equal to, the equality doesn't hold and therefore inf (S) < sup (S) as it can never be greater than it. We have reached a contradiction, and this statement is false.

2

Using the information in the above (since I don't want to make it really long), if the set only contains one element, then the smallest and the largest element will either be the same, or it will have a smallest element and tends towards a largest element, or it will have a largest element and tends towards a smallest element unsure about these last two.Again, this whole nonsense about "tending towards" is messing you up. Use the actual definitions for sup and inf and this is trivial..

Is this wrong? E.g. of it will have a smallest element and tends towards a largest element: Let x = {X is an integer | 1 ≤ x < 2}. It doesn't tend towards 2 right because x is not a real number, it is an integer? So would this be incorrect?Integers are real numbers!.

If it is incorrect, I'll just prove the first statement I don't actually know what you mean by "first statement" here; when you say something like this you should make sure it actually *is* obvious what statement you're talking about: a set where the smallest and the largest element will be the same e.g. a. This isn't a proper sentence (it's the mathematical equivalent of "a red apple").
In this case, any element ≤ a or ≥ a will be a lower bound and upper bound respectively. So, inf (S) = sup (S) = a.
In this case, the inequality holds and therefore, the strict inequality isn't satisfied - we have reached a contradiction and therefore this statement is false.

As both statements are false, this leads us to conclude that the original statement is true and therefore inf S ≤ sup S, and that the equality holds iff S is a singleton set (that is a set with exactly one element).

By the way, I know you'll be really tired by the time you read all of that, so if you have time, I'd really appreciate if you can just confirm the next part:


Contrapositive:


I assume what I've written below is the contrapositive, can you please confirm that?

For any non-empty bounded set S, prove that S is not a singleton set (that is a set with exactly one element) iff inf S > sup S, and that the equality doesn't hold, only in this case. It's never the case that inf S > sup S. Did you mean inf S < sup S?

Ignoring that, iff is really 2 statements, so you should treat them both seperately, not try to take the contrapositive of their combination

Would this be easier to prove in any way, or is it just the same approach but sort of reversed (like backwards) i.e. proving that if inf S > sup S, then S is not a singleton set ... etc."the same approach but sort of reversed" is exactly what the contrapositive is. .
It should perhaps be said that you should be able to answer this in 3-4 lines (a bit more if you really wanted to spell out every detail). So as soon as you start looking at writing half a page for one part of it, you should probably think again.
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Chittesh14
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(Original post by ghostwalker)
Haven't got the energy at present to even fully read through that, so a general statement:

Iff statements are usually proved by proving the logically equaivalant "(A impiles B) AND (B implies A)" and each part is treated separately.

It is often the case that doing the implication one way is straight forward, almost trivial. And doing the implication the other way is somewhat harder and this part is often best done via proof by contradiction.

All that, rather than negating the original iff in using proof by contradtion on the whole thing.
That's fine, thank you. This is the first time I have seen an iff implication, so wasn't sure as to what to do. Thank you .
(Original post by DFranklin)
I agree with Ghostwalkers advice, but I think some comments on what you've done are necessary. I've commented inline, in red. It should be said that a lot of what I haven't commented on isn't really "right" either, but there's so many errors that often your subsequent work just doesn't make sense.



It should perhaps be said that you should be able to answer this in 3-4 lines (a bit more if you really wanted to spell out every detail). So as soon as you start looking at writing half a page for one part of it, you should probably think again.
Thank you . I thought I misinterpreted the question! lol

What's all this "set tends towards" nonsense? There's a very informal way in which it might make sense, but I can't believe it's any part of your formal definition of inf(S) and sup(S) and it's certainly too informal to be applicable here.

The only problem here is that, I don't know how to define it. We've been told that the infimum is the greatest lower bound i.e. the highest value of x such that x ≤ (Every member in the set).
But, sometimes the member is included in the set and sometimes it is not (I've seen in examples), but we haven't ever been shown a way to prove using infimum, just to work it out. So I was finding it difficult how to prove it and justify it like you said, I could only write down what I mean and I understand it is very confusing, so I'm sorry about that .

E.g. of it will have a smallest element and tends towards a largest element: Let x = {X is an integer | 1 ≤ x < 2}. It doesn't tend towards 2 right because x is not a real number, it is an integer?

Sorry, yeah I know integers are real numbers, but would the upper bound of this set be x ≥ 1 because 1 is the only element (i.e. it is a singleton set)?

For any non-empty bounded set S, prove that S is not a singleton set (that is a set with exactly one element) iff inf S > sup S, and that the equality doesn't hold, only in this case. It's never the case that inf S > sup S. Did you mean inf S < sup S?

Not sure here .

Thank you so much for your help, I think I just need to have another attempt at this question and rethink what I've done (after misinterpreting the question) and everything. I'll try to make a better proof next time which makes more sense .
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DFranklin
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(Original post by Chittesh14)
The only problem here is that, I don't know how to define it. We've been told that the infimum is the greatest lower bound i.e. the highest value of x such that x ≤ (Every member in the set).
Yes, that's the definition. That's what you need to work with. None of this "set tends to..." nonsense.

But, sometimes the member is included in the set and sometimes it is not (I've seen in examples), but we haven't ever been shown a way to prove using infimum, just to work it out. So I was finding it difficult how to prove it and justify it like you said, I could only write down what I mean and I understand it is very confusing, so I'm sorry about that .
The thing is, every time you write "tends to" (in this discussion), you're basically writing nonsense. And it's very hard to know what you actually mean, when what you're writing is nonsense.

There are two things you know about the infinum. It's a lower bound, and there can't be a bigger lower bound.

A particular observation that will help you here: if a \in S, then \inf S \leq a (because otherwise inf S isn't a lower bound).

[/b]Sorry, yeah I know integers are real numbers, but would the upper bound of this set be x ≥ 1 because 1 is the only element (i.e. it is a singleton set)?
"x ≥ 1" is a statement about x, not an upper bound. I honestly don't know what you're trying to say here - I mean "x ≥ 1" is also a statement that implies 1 is a lower bound and possibly you meant *that*, but it tells you absolutely nothing about the upper bound.

For any non-empty bounded set S, prove that S is not a singleton set (that is a set with exactly one element) iff inf S > sup S, and that the equality doesn't hold, only in this case. It's never the case that inf S > sup S. Did you mean inf S < sup S?

Not sure here .
In that case, you honestly need to start again, using the approach ghostwalker (and I) have suggested.

[There are many things that *look* like "silly mistakes" in what you've written, like saying inf S > sup S, but also, say, "negating" a statement like a \leq b to get a \geq b (where the correct negation is a &gt; b). But you've so over-complicated things that it's hard to be sure which are silly (easily corrected) mistakes and which are fundamental logical errors].
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Chittesh14
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(Original post by DFranklin)
Yes, that's the definition. That's what you need to work with. None of this "set tends to..." nonsense.

The thing is, every time you write "tends to" (in this discussion), you're basically writing nonsense. And it's very hard to know what you actually mean, when what you're writing is nonsense.

There are two things you know about the infinum. It's a lower bound, and there can't be a bigger lower bound.

A particular observation that will help you here: if a \in S, then \inf S \leq a (because otherwise inf S isn't a lower bound).

"x ≥ 1" is a statement about x, not an upper bound. I honestly don't know what you're trying to say here - I mean "x ≥ 1" is also a statement that implies 1 is a lower bound and possibly you meant *that*, but it tells you absolutely nothing about the upper bound.

In that case, you honestly need to start again, using the approach ghostwalker (and I) have suggested.

[There are many things that *look* like "silly mistakes" in what you've written, like saying inf S > sup S, but also, say, "negating" a statement like a \leq b to get a \geq b (where the correct negation is a &gt; b). But you've so over-complicated things that it's hard to be sure which are silly (easily corrected) mistakes and which are fundamental logical errors].
Thanks, I'll just start again I guess lol. I think I'll simplify it now so they'll be less mistakes. Yeah, most of those errors are because I just find it hard to type on TSR and often copy and paste symbols. Sorry about that.
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Chittesh14
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(Original post by DFranklin)
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(Original post by ghostwalker)
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Hello,

Just wondering if the statement is as follows: For every non-empty bounded set S, if inf S < x < sup S, then x ∈ S.

If I wanted to prove this wrong, i.e. negation is true. Do I have to prove that this statement below is true:

There exists a non-empty bounded set S, such that, inf S < x < sup S and x ∉ A?

Thanks for your help .

BTW is this a correct proof (I think the statement is original true):

As inf S ≤ x for all x ∈ S. Then, either inf S < x for all x ∈ S or inf S = x for all x ∈ S. I get a bit confused here the way I write it, because sometimes I feel I'm writing inf S has to be equal to every x in the set, which is impossible and is not what I mean to say As a result, for all x ∈ S, inf S can be either: < x or = x), then if inf S < x, then x ∈ S.
Similarly, sup S ≥ x for all x ∈ S. As a result, for all x ∈ S, sup S > x or sup S = x. So, if sup S > x, then x ∈ S.

As a result, combining the two statements together: if inf S < x, then x ∈ S and So, if sup S > x, then x ∈ S.
If inf S < x and sup S > x, x ∈ S.
So, if inf S < x < sup S, then x ∈ S.

Is this correct?
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(Original post by Chittesh14)
Hello,

Just wondering if the statement is as follows: For every non-empty bounded set S, if inf S < x < sup S, then x ∈ S.
I'm not sure why you think this is relevant, but I am sure that it's false. Take S ={1,3} then inf S = 1, sup S = 3 and 2 \notin S
If I wanted to prove this wrong, i.e. negation is true. Do I have to prove that this statement below is true:

There exists a non-empty bounded set S, such that, inf S < x < sup S and x ∉ A?
You mean x ∉ S. And you need to insert something like "we can find x with" after "such that".


BTW is this a correct proof (I think the statement is original true):

As inf S ≤ x for all x ∈ S. Then, either inf S < x for all x ∈ S or inf S = x for all x ∈ S. I get a bit confused here the way I write it, because sometimes I feel I'm writing inf S has to be equal to every x in the set, which is impossible and is not what I mean to say
Well, that *is* what your saying. If you don't mean to say that, don't say it...

As result, for all x ∈ S, inf S can be either: < x or = x), then if inf S < x, then x ∈ S.
The part in bold isn't true. To be blunt, it should be obvious that it's not true. Again, consider the counterexample set that I posted.

Similarly, sup S ≥ x for all x ∈ S. As a result, for all x ∈ S, sup S > x or sup S = x. So, if sup S > x, then x ∈ S.

As a result, combining the two statements together: if inf S < x, then x ∈ S and So, if sup S > x, then x ∈ S.
If inf S < x and sup S > x, x ∈ S.
So, if inf S < x < sup S, then x ∈ S.

Is this correct?
No, it's not.
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(Original post by DFranklin)
I'm not sure why you think this is relevant, but I am sure that it's false. Take S ={1,3} then inf S = 1, sup S = 3 and 2 \notin S
You mean x ∉ S. And you need to insert something like "we can find x with" after "such that".

Well, that *is* what your saying. If you don't mean to say that, don't say it...

The part in bold isn't true. To be blunt, it should be obvious that it's not true. Again, consider the counterexample set that I posted.

No, it's not.
Thank you, I've got it.
Damn, that was really stupid of me lol, I think I've just been dealing with continuous sets, so I've sort of forgot they can be discrete too!

Thank you so much
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(Original post by DFranklin)
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But let's say the set was continuous e.g. A = {x is a real number | a < x < b}, then if inf S < x, then x ∈ S, right?
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(Original post by Chittesh14)
But let's say the set was continuous e.g. A = {x is a real number | a < x < b}, then if inf S < x, then x ∈ S, right?
Technically, I think what you mean is "connected", not continuous. And you're mixing up A and S again. But assuming you meant S throughout - it's still not true: inf S < b but b is not in S.

But focussing on sets of this form won't help you answer the problem anyhow.
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(Original post by Chittesh14)
Thanks, I'll just start again I guess lol. I think I'll simplify it now so they'll be less mistakes. Yeah, most of those errors are because I just find it hard to type on TSR and often copy and paste symbols. Sorry about that.
Because it doesn't feel this is going anywhere, I'm going to give you a framework to use to answer the (iff part of this) question.

(Claim 1): If S has only one element, so that S = {a}, say, then inf S = sup S.
(1a) Hopefully you have a pretty good idea what inf S and sup S must be, so you just need to prove that that number is an upper and lower bound and that there's no smaller upper bound and bigger lower bound.

(Claim 2) Prove that is S has more than one element, then inf S < sup S.
(2a) If S has more than one element, that it has at least two elements.
(2b) So you can pick distinct a, b in S.
(2c) Now that you have 2 distinct elements, what can we say about inf S and sup S? (in relation to a and b)
(2d) Deduce that inf S < sup S (and equality cannot hold).

(Conclusion): Note that (by negation), proving claim 2 this will prove that if inf S >= sup S, then S has only one element. Which implies that if inf S = sup S, then S has only one element. So proving (1) and (2) is enough to show that for non-empty S, inf S = sup S iff S has exactly one element.

Use this framework and see how you get on.
Notes: Try to follow the steps exactly instead of skipping around and meandering all over the place. I don't think the answer to any step should be longer than a sentence or 2, with the possible exception of (1a). And some steps (2a, 2b, for example) don't really require any justification at all.

I'm not sure if you managed to prove inf S <= sup S, but if you didn't, just pretend you did for now.
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(Original post by DFranklin)
Technically, I think what you mean is "connected", not continuous. And you're mixing up A and S again. But assuming you meant S throughout - it's still not true: inf S < b but b is not in S.

But focussing on sets of this form won't help you answer the problem anyhow.
Sorry, it was early morning lol. Idk why I keep mixing it up, I'm really sorry. Ok yeah, I get what you mean. Like S = {x is real | 1 < x < 5}, inf(S) = 1 and b could be not in S e.g. 10 and inf(S) < b.
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(Original post by DFranklin)
Because it doesn't feel this is going anywhere, I'm going to give you a framework to use to answer the (iff part of this) question.

(Claim 1): If S has only one element, so that S = {a}, say, then inf S = sup S.
(1a) Hopefully you have a pretty good idea what inf S and sup S must be, so you just need to prove that that number is an upper and lower bound and that there's no smaller upper bound and bigger lower bound.

(Claim 2) Prove that is S has more than one element, then inf S < sup S.
(2a) If S has more than one element, that it has at least two elements.
(2b) So you can pick distinct a, b in S.
(2c) Now that you have 2 distinct elements, what can we say about inf S and sup S? (in relation to a and b)
(2d) Deduce that inf S < sup S (and equality cannot hold).

(Conclusion): Note that (by negation), proving claim 2 this will prove that if inf S >= sup S, then S has only one element. Which implies that if inf S = sup S, then S has only one element. So proving (1) and (2) is enough to show that for non-empty S, inf S = sup S iff S has exactly one element.

Use this framework and see how you get on.
Notes: Try to follow the steps exactly instead of skipping around and meandering all over the place. I don't think the answer to any step should be longer than a sentence or 2, with the possible exception of (1a). And some steps (2a, 2b, for example) don't really require any justification at all.

I'm not sure if you managed to prove inf S <= sup S, but if you didn't, just pretend you did for now.
Sorry, I actually done this last night but didn't post it.

Posting what I originally done (without reading this), then a next post will be an update after reading this:

1: For any non-empty bounded set S, inf S ≤ sup S.

If S is bounded and non-empty, then there will be a supremum and infimum of the set.
As inf S ≤ x for all x ∈ S and sup S ≥ x for all x ∈ S.
So inf S ≤ x ≤ sup S and therefore inf S ≤ sup S.

I really hope that is correct!

2. If inf S = sup S, then S is a singleton set.

Proof by contradiction:

inf S = sup S and S is not a singleton set.

If S in not a singleton set, it has more than 1 element of different values.
So, there will be one element greater than all the other elements (since they're all distinct). Let that be G.
There will be one element less than all the other elements, let that be L.

In this case, as the elements are not the same i.e. L ≠ G and G > L.
Using the definition, inf S ≤ L and sup S ≥ G, where L ≠ G.
So, inf S ≤ L < G and sup S ≥ G > L and so inf S < G and sup S ≥ G.
Hence, inf S ≠ sup S.

So, we have reached a contradiction - inf S = sup S and also inf S ≠ sup S.
So, the original statement is true - if inf S = sup S, then S is a singleton set.

3. If S is a singleton set, then inf S = sup S.

If S in a singleton set, it only has one element. Let the element be x.
So inf S ≤ x and sup S ≥ x and so inf S ≤ x ≤ sup S.
So, inf S ≤ sup S.
But, as x is the only element in the set, the greatest element ≤ to x, and the smallest element ≥ to x, both will be x itself.
Therefore, inf S = x and sup S = x and therefore inf S = sup S.
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Chittesh14
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(Original post by DFranklin)
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(Claim 1): If S has only one element, so that S = {a}, say, then inf S = sup S.
(1a) Hopefully you have a pretty good idea what inf S and sup S must be, so you just need to prove that that number is an upper and lower bound and that there's no smaller upper bound and bigger lower bound.

Proof: If S has only element, e.g. S = {a}, then inf S = sup S.
inf S ≤ a and sup S ≥ a. In fact, any lower bound (l) will be of the form: l ≤ a and similarly any upper bound (u) will be of the form: u ≥ a.
The greatest lower bound would be when l = a and the smallest upper bound would be when u = a.
Not exactly sure how to show there's no smaller upper bound and bigger lower bound .


(Claim 2) Prove that is S has more than one element, then inf S < sup S.
(2a) If S has more than one element, that it has at least two elements.
(2b) So you can pick distinct a, b in S.
(2c) Now that you have 2 distinct elements, what can we say about inf S and sup S? (in relation to a and b)
(2d) Deduce that inf S < sup S (and equality cannot hold).

Proof: If S has more than one element, it has at least two elements. Let the two distinct elements be a and b. Assume b is the greater out of the two, therefore a < b. So, inf S ≤ a and sup S ≥ b. So, inf S ≤ a < b and sup S ≥ b > a. So inf S ≤ a and sup S > a. So, inf S < sup S. The equality cannot hold.

(Claim 3): Prove that for any non-empty bounded set S, inf S ≤ sup S.

Proof: As inf S ≤ x for all x ∈ S and sup S ≥ x for all x ∈ S. So inf S ≤ x ≤ sup S and therefore inf S ≤ sup S.

(Conclusion): By negation, proving claim 2 this will prove that if inf S >= sup S, then S has only one element. Which implies that if inf S = sup S, then S has only one element. So proving (1) and (2) is enough to show that for non-empty S, inf S = sup S iff S has exactly one element.

Does this mean I can just prove (Claim 1) like this: Since I have proven that if S has more than 1 element, then inf S < sup S. Then, the contrapositive will be: If inf S >= sup S, then S only has one element. Hence, the equality (i.e. inf S = sup S) only holds when S has one element.
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DFranklin
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(Original post by Chittesh14)
Sorry, I actually done this last night but didn't post it.

Posting what I originally done (without reading this), then a next post will be an update after reading this:

1: For any non-empty bounded set S, inf S ≤ sup S.

If S is bounded and non-empty, then there will be a supremum and infimum of the set.
As inf S ≤ x for all x ∈ S and sup S ≥ x for all x ∈ S.
So inf S ≤ x ≤ sup S and therefore inf S ≤ sup S.
This is kind of OK, but I'm pretty sure it's more by luck than anything else - you've been inconsistent about saying whether you mean "for all x" or not. It would be cleaner to just pick any x in S, you still have inf S <= x < = sup S.

I really hope that is correct!

2. If inf S = sup S, then S is a singleton set.

Proof by contradiction:

inf S = sup S and S is not a singleton set.

If S in not a singleton set, it has more than 1 element of different values.
So, there will be one element greater than all the other elements (since they're all distinct). Let that be G.
NO! This is a really fundamental error (albeit one easily made). The whole point of introducing the concepts of sup and inf is that you cannot guarantee S has such an element. For example if S is the set {x: -1 < x < 2}, then there's no such element G in S. (You would want G = 2, but 2 is not an element of S).

3. If S is a singleton set, then inf S = sup S.

If S in a singleton set, it only has one element. Let the element be x.
So inf S ≤ x and sup S ≥ x and so inf S ≤ x ≤ sup S.
So, inf S ≤ sup S.
But, as x is the only element in the set, the greatest element ≤ to x, and the smallest element ≥ to x, both will be x itself.
Therefore, inf S = x and sup S = x and therefore inf S = sup S.
This is kind of OK, but the definition of inf S talks about "greatest upper bound", and you haven't talked about bounds at all.
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