I'm not sure if my working is right for this one..
I have put:
For all epsilon > 0, there exists delta > 0, such that  f(x)  f(a)  < epsilon,  x  a  < delta
We can use a delta small enough to take epsilon = f(a)/2
 f(x)  f(a)  =< f(a)/2 = epsilon
Resulting in 3f(a)/2 > f(x) > f(a)/2
But I'm not sure if that's okay?
Next question:
Let x >= z, and let f(z) = inf{x : x E R}
Let f(z) = c
Take a small delta such that  x  z  < delta
Then I went to choose c such that f(x)  c < f(x)  1 < epsilon
Again, I feel like I'm not doing well enough here.
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TheBBQ
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 15012015 22:15

ghostwalker
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 15012015 23:34
(Original post by TheBBQ)
I'm not sure if my working is right for this one..
I have put:
For all epsilon > 0, there exists delta > 0, such that  f(x)  f(a)  < epsilon,  x  a  < delta
We can use a delta small enough to take epsilon = f(a)/2
 f(x)  f(a)  =< f(a)/2 = epsilon
Resulting in 3f(a)/2 > f(x) > f(a)/2
But I'm not sure if that's okay?
Your definition of continuity is a bit sloppy.
We choose epsilon, f(a)/2, then there exists a delta such that if  xa  < delta, then f(x)  f(a)  < f(a)/2.
And continue.
Also, take care to use strict inequalities "<", or weak inequalities "<=", correctly. Sometimes it's not a significant part of a proof, but on other occasions the whole proof stands or falls on it.
Next question:
Let x >= z, and let f(z) = inf{x : x E R}
Let f(z) = c
Take a small delta such that  x  z  < delta
Then I went to choose c such that f(x)  c < f(x)  1 < epsilon
Again, I feel like I'm not doing well enough here. 
TheBBQ
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 15012015 23:41
(Original post by ghostwalker)
Basically OK, but needs sharpening up a bit.
Your definition of continuity is a bit sloppy.
We choose epsilon, f(a)/2, then there exists a delta such that if  xa  < delta, then f(x)  f(a)  < f(a)/2.
And continue.
Also, take care to use strict inequalities "<", or weak inequalities "<=", correctly. Sometimes it's not a significant part of a proof, but on other occasions the whole proof stands or falls on it.
What's z? and your set "inf{x : x E R}" doesn't make much sense.
Yeah I'm thinking that I should use weak inequalities for most of the first question, and then finally saying 3f(a)/2 >= f(x) >= f(a)/2, and then hence f(x) >= f(a)/2?
For the second part, I meant to say that I'm using the extreme value theorem, taking z in [0, infinity) such that all x >= z, hence f(z) = {f(x): x E R), I acidentally typed it up a bit wrongly 
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 15012015 23:47
(Original post by TheBBQ)
Thanks for the help
Yeah I'm thinking that I should use weak inequalities for most of the first question, and then finally saying 3f(a)/2 >= f(x) >= f(a)/2, and then hence f(x) >= f(a)/2?
For the second part, I meant to say that I'm using the extreme value theorem, taking z in [0, infinity) such that all x >= z, hence f(z) = {f(x): x E R), I acidentally typed it up a bit wrongly
Also make sure you meet the criteria for the extreme value theorem in your usage of it. 
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 15012015 23:53
(Original post by ghostwalker)
Well the question is asking for you to show f(x) > f(a)/2
Can you post your answer to the second question all together, as I'm not clear what's going on trying to combine that with what you said in your first post.
Also make sure you meet the criteria for the extreme value theorem in your usage of it.
Here is my answer for the second part:
Take z in [0, infinity)
Let x >= z, and let f(z) = inf{f(x) : x E R} I'm not sure if this should be x in R or x in [0, infinity)..
Let f(z) = c, which we can do as f(z) is in the infimum of function and it is required to find f(x) >= c for all x in [0,infinity)
Take a small delta such that  x  z  < delta
Then I went to choose c such that f(x)  c =< f(x)  1 =< epsilon
That is my correction. 
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 16012015 00:03
(Original post by TheBBQ)
But is that way of doing the first question acceptable?
Here is my answer for the second part:
Take z in [0, infinity)
Let x >= z, and let f(z) = inf{f(x) : x E R} I'm not sure if this should be x in R or x in [0, infinity)..
Last post for today. 
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 16012015 00:24
(Original post by ghostwalker)
In essence yes, with the caveats I previously mentioned. You are asked to show f(x) > f(a)/2, not >= f(a)/2, the latter being a weaker condition.
OK, I choose z=5
But f(x) is defined already, in particular f(z) = f(5), so I can't define it again with that set.
Last post for today.
Hmm I see that what I have done wouldn't work, but I am rather unsure of the correct way to go.
I'll try again, hopefully someone else can help
As f is continous, we can find c > 0 such that f(x) >= c > 0
For f(x) >= c to hold true for every x E [0,infinity), let x >= z, and let f(z) = inf{f(x) : x E [0,infinity) }
Take a small delta such that  x  z  < delta
Then I went to choose c such that f(x)  c =< f(x)  1 =< epsilon 
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 16012015 02:17
(Original post by TheBBQ)
..
There are 2 separate things going on here  you've told f(x) is always > 0, and you're also told that f(x)>1 as x>infinity.
Now on any finite interval I, you can use the extreme value theorem to say f(x) attains it's bounds on I, and so since f(x) > 0, the bound on I is also > 0.
But your problem is that is NOT a finite interval.
So you're going to need to use what you're told about the behaviour of f(x) as x>infinity.
I would start by writiing down what it means to say that as . Then use that to show that you can find an infinite interval on which we know f(x) > 1/2. Then extend to the full range by adding on a finite interval. 
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 16012015 10:11
(Original post by DFranklin)
To my mind it's going to be very difficult to solve the 2nd question using just one idea (as it seems you're currently trying to do).
There are 2 separate things going on here  you've told f(x) is always > 0, and you're also told that f(x)>1 as x>infinity.
Now on any finite interval I, you can use the extreme value theorem to say f(x) attains it's bounds on I, and so since f(x) > 0, the bound on I is also > 0.
But your problem is that is NOT a finite interval.
So you're going to need to use what you're told about the behaviour of f(x) as x>infinity.
I would start by writiing down what it means to say that as . Then use that to show that you can find an infinite interval on which we know f(x) > 1/2. Then extend to the full range by adding on a finite interval.
Taking epsilon = 1/2, and using an M in R, and letting x >= M, then I can have it such that  f(x)  1  =< 1/2 = epsilon, and from that we have f(x) >= 1/2
Now I can take the other part of the interval, [0.M], and find c in here, which would mean that f(x) >= 1/2 >= c > 0
I can find a z in the interval [0,M] such that f(z) = inf { f(m): m E [0,M] } > 0, and call this c
Then it would satisfy that f(x) >= 1/2 >= c as sup { f(m): m E [0,M] } would be =< 1/2?Last edited by TheBBQ; 16012015 at 10:23. 
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 16012015 10:59
(Original post by TheBBQ)
I think f(x) > 1 as x > infinity means f(x) is continous on all points between 0 and infinity, thus we can find c > 0
as if such that .
Taking epsilon = 1/2, and using an M in R, and letting x >= M, then I can have it such that  f(x)  1  =< 1/2 = epsilon, and from that we have f(x) >= 1/2
Now I can take the other part of the interval, [0.M], and find c in here, which would mean that f(x) >= 1/2 >= c > 0
I can find a z in the interval [0,M] such that f(z) = inf { f(m): m E [0,M] } > 0, and call this c
Then it would satisfy that f(x) >= 1/2 >= c as sup { f(m): m E [0,M] } would be =< 1/2? 
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 16012015 12:20
(Original post by DFranklin)
No, this isn't what it means. There is no implication of continuity. A correct definition:
as if such that .
This is basically correct, but it's very unclearly worded.
Not sure what you're trying to say here. Some of this isn't true, and what is true needs justification, which you haven't provided. I think the justification is supposed to be what's following, but you either need to put the justification first or rewrite the above to be more like "We still need to deal with the interval [0,M]. We shall show that we can find c > 0 such that ..."
If you're going to use a theorem, you should actually name it, and make it clear that any conditions are satisfied. e.g. "Since f is cts, and [0, M] is a closed interval, f attains its bounds on this interval. In particular, we can find z in [0, M] such that..."
There is no reason to assume c <= 1/2. E.g. f(x) = 1 everywhere (f is constant). You also seem to have switched from inf to sup but I assume that's a typo.
As f is continous in all of [0,infinity), by the extreme value theorem we can find a z in the finite interval [0,M] such that f(z) = sup { f(m): m E [0,M] } > 0, and denote this as c, and we have that c > 0.
As we have f(x) >= 1/2 as x > infinity in the interval [M,infinity) , it stands that 1/2 >= f(z) = c and thus f(x) >= 1/2 >= c? This is all i can think of
I am not sure how I am meant to do this last bit, as I have found that f(x) >= 1/2, but I don't think I can actually justify using c =< 1/2 besides that I've made it the sup of the finite interval..Last edited by TheBBQ; 16012015 at 12:22. 
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 16012015 13:00
(Original post by TheBBQ)
Yes that last bit is the typo.
As f is continous in all of [0,infinity), by the extreme value theorem we can find a z in the finite interval [0,M] such that f(z) = sup { f(m): m E [0,M] } > 0, and denote this as c, and we have that c > 0.
Previously I posted:
(Original post by DFranklin)
I would start by writiing down what it means to say that f(x)\to 1 as x \to \infty. Then use that to show that you can find an infinite interval [M, \infty) on which we know f(x) > 1/2. Then extend to the full range [0,\infty) by adding on a finite interval.
[I must say, presenting your argument in the right order seems to be a bit of a problem for you  you keep using variables in paragraph 1 that you don't define until paragraph 3 etc. You need to sort this out].
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