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Analysis - Open sets and continuous functions

Well at this point I'm really struggling with Analysis :lol:

Capture.PNG

Not quite sure how to go about the whole thing - starting with the first proof.

To begin with I assume that there exists an open set with finitely many elements in Ω\Omega then straight from the definition of an open set I know that ∀x∈Ω,∃ϵ>0,s.t.Bϵ(x)⊆Ω\forall x \in \Omega, \quad \exists \epsilon > 0, \quad s.t.\quad B_{\epsilon}(x)\subseteq \Omega but I'm unsure how to continue on to show that there are infinitely many x∈Ωx \in \Omega via contradiction
(edited 7 years ago)

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Original post by RDKGames
Well at this point I'm really struggling with Analysis :lol:

Capture.PNG

Not quite sure how to go about the whole thing - starting with the first proof.

To begin with I assume that there exists an open set with finitely many elements in Ω\Omega then straight from the definition of an open set I know that ∀x∈Ω,∃ϵ>0,s.t.Bϵ(x)⊆Ω\forall x \in \Omega, \quad \exists \epsilon > 0, \quad s.t.\quad B_{\epsilon}(x)\subseteq \Omega but I'm unsure how to continue on to show that there are infinitely many x∈Ωx \in \Omega via contradiction


Caveat: My analysis is rusty.

How many elements of Rn\mathbb{R}^n are in Bϵ(x)B_{\epsilon}(x) ?
(edited 7 years ago)
Original post by RDKGames
Well at this point I'm really struggling with Analysis :lol:

Capture.PNG

Not quite sure how to go about the whole thing - starting with the first proof.

To begin with I assume that there exists an open set with finitely many elements in Ω\Omega then straight from the definition of an open set I know that ∀x∈Ω,∃ϵ>0,s.t.Bϵ(x)⊆Ω\forall x \in \Omega, \quad \exists \epsilon > 0, \quad s.t.\quad B_{\epsilon}(x)\subseteq \Omega but I'm unsure how to continue on to show that there are infinitely many x∈Ωx \in \Omega via contradiction


Let Ω={xi∈Rn:0≤i≤k,i,k∈N}\Omega = \{ x_i \in \mathbb{R}^n : 0 \le i \le k, i,k \in \mathbb{N} \}, so that Ω\Omega is finite.

Now choose xi∈Ωx_i \in \Omega, and we know that since Ω\Omega is open, we must have Bϵ(xi)⊆ΩB_{\epsilon}(x_i)\subseteq \Omega for a suitable ϵ\epsilon. Now if you can show that there must be some xjx_j that is not in the enumeration of the set given above, then you have a contradiction.

To do this, I think that you want to consider the set of distances between the xix_i.
Original post by atsruser
Let Ω={xi∈Rn:0≤i≤k,i,k∈N}\Omega = \{ x_i \in \mathbb{R}^n : 0 \le i \le k, i,k \in \mathbb{N} \}, so that Ω\Omega is finite.

Now choose xi∈Ωx_i \in \Omega, and we know that since Ω\Omega is open, we must have Bϵ(xi)⊆ΩB_{\epsilon}(x_i)\subseteq \Omega for a suitable ϵ\epsilon. Now if you can show that there must be some xjx_j that is not in the enumeration of the set given above, then you have a contradiction.

To do this, I think that you want to consider the set of distances between the xix_i.
I don't see any of this is necessary - Ghostwalker's post is sufficient.

[Your approach is better suited to showing we can find a ball that is completely disjoint from the x_i, which is a stronger result. Its probably a good way to show the set is *closed*, however]
Reply 4
Avatar for math42
math42
20
Well it is sort of obvious that the ball of radius epsilon centred at x has infinitely many elements. If you want to be on the safe side, it is easy to find infinitely many, you can just find a bunch of vectors with the same component as x in the 2nd to nth places and then put some sequence that tends to x1 for the first components
Original post by ghostwalker
Caveat: My analysis is rusty.

How many elements of Rn\mathbb{R}^n are in Bϵ(x)B_{\epsilon}(x) ?


Coming from the assumption, there are finitely many so we can say there's k∈Nk\in \mathbb{N} of them as shown in the question, not sure how to use this fact.
(edited 7 years ago)
Original post by RDKGames
Coming from the assumption, there are finitely many so we can say there's k∈Nk\in \mathbb{N} of them as shown in the question, not sure how to use this fact.


Just to be clear, my response is dealing with the very first part of your original question.

You've shown that Bϵ(x)B_{\epsilon}(x) lies within your set Ω\Omega. But this is an open ball in Rn\mathbb{R}^n so it has an uncountably infinite number of elements. Hence so does Ω\Omega - contradiction.
Original post by ghostwalker
Just to be clear, my response is dealing with the very first part of your original question.

You've shown that Bϵ(x)B_{\epsilon}(x) lies within your set Ω\Omega. But this is an open ball in Rn\mathbb{R}^n so it has an uncountably infinite number of elements. Hence so does Ω\Omega - contradiction.


Oh I just realised I may have proven before that an open ball has infinitely many elements - makes the whole question seem so much easier now with this... Thanks!

Now for the second part, a similar construction can be made here I think? Say A⊆RnA \subseteq \mathbb{R}^n is finite and open then by the same contradiction it must be closed??
(edited 7 years ago)
Original post by RDKGames

Now for the second part, a similar construction can be made here I think? Say A⊆RnA \subseteq \mathbb{R}^n is finite and open then by the same contradiction it must be closed??


If a subset isn't open, it does not necessarily imply that it is closed.

E.g. consider the interval (0,1] in R. It is neither open, nor closed.


To show that a finite subset of Rn\mathbb{R}^n is closed, I'd look to show that its complement in Rn\mathbb{R}^n is open.

Big hint in spoiler.

Spoiler

(edited 7 years ago)
Original post by ghostwalker
If a subset isn't open, it does not necessarily imply that it is closed.

E.g. consider the interval (0,1] in R. It is neither open, nor closed.


To show that a finite subset of Rn\mathbb{R}^n is closed, I'd look to show that its complement in Rn\mathbb{R}^n is open.

Big hint in spoiler.

Spoiler




So if Ω={xi∈Rn:0≤i≤k,i,k∈N}⊆Rn\Omega = \{ x_i \in \mathbb{R}^n : 0 \le i \le k, i,k \in \mathbb{N} \} \subseteq \mathbb{R}^n then if I choose ϵ=∣xi−c∣\epsilon = \vert x_i -c \vert where c∈Rn\Ωc\in \mathbb{R}^n \backslash \Omega it gives out Bϵ(xi)⊆Rn\ΩB_{\epsilon}(x_i) \subseteq \mathbb{R}^n \backslash \Omega thus Rn\Ω\mathbb{R}^n\backslash \Omega is open hence Ω\Omega (the complement) is closed as required.

Does that sound right?? I'm not fully sure about the middle of that bit but the end makes sense to me.
(edited 7 years ago)
Reply 10
Avatar for Zacken
Zacken
22
A different approach to the above: if you have the result that a singleton set is closed (not hard to prove: R^n \ {x} is open) and that a finite union of closed sets is closed then you're done.

Edit: that said, it's still pretty much the same argument at the end of the day. Just repackaged a little.
(edited 7 years ago)
Original post by RDKGames
So if Ω={xi∈Rn:0≤i≤k,i,k∈N}⊆Rn\Omega = \{ x_i \in \mathbb{R}^n : 0 \le i \le k, i,k \in \mathbb{N} \} \subseteq \mathbb{R}^n then if I choose ϵ=∣xi−c∣\epsilon = \vert x_i -c \vert where c∈Rn\Ωc\in \mathbb{R}^n \backslash \Omega it gives out Bϵ(xi)⊆Rn\ΩB_{\epsilon}(x_i) \subseteq \mathbb{R}^n \backslash \Omega thus Rn\Ω\mathbb{R}^n\backslash \Omega is open hence Ω\Omega (the complement) is closed as required.

Does that sound right?? I'm not fully sure about the middle of that bit but the end makes sense to me.


Zacken's (PRSOM) suggestion - above - would be quicker. It all comes down to what you've covered already on your course.

Going back to your response:

It's nearly there. Your choice of epsilon seems to depend on just one of the xix_i. For a given c (defined as per your post), you need to choose epsilon such that your open ball does not include any element of Ω\Omega, so just tighten that up a bit.

Also you're trying to show Bϵ(c)⊆Rn\ΩB_{\epsilon}(c) \subseteq \mathbb{R}^n \backslash \Omega, not x_i, since c is your chosen arbitrary point in Rn\Ω\mathbb{R}^n \backslash \Omega - just a typo, I assume.
(edited 7 years ago)
Original post by ghostwalker
Zacken's (PRSOM) suggestion - above - would be quicker. It all comes down to what you've covered already on your course.

Going back to your response:

It's nearly there. Your choice of epsilon seems to depend on just one of the xix_i. For a given c (defined as per your post), you need to choose epsilon such that your open ball does not include any element of Ω\Omega, so just tighten that up a bit.

Also you're trying to show Bϵ(c)⊆Rn\ΩB_{\epsilon}(c) \subseteq \mathbb{R}^n \backslash \Omega, not x_i, since c is your chosen arbitrary point in Rn\Ω\mathbb{R}^n \backslash \Omega - just a typo, I assume.


I'm taking an approach with a diagram for help and so I can see that for the open ball Bϵ(c)⊆Rn\ΩB_{\epsilon}(c) \subseteq \mathbb{R}^n \backslash \Omega to not include any element of Ω\Omega, it must be true that my choice for epsilon must instead be defined as ϵ=min⁡0≤i≤k∣xi−c∣\displaystyle \epsilon = \min_{0\leq i \leq k}\vert x_i-c \vert ??

P.S. Zacken's response makes sense, but I think we are supposed to take the approach concerning distances :smile:
(edited 7 years ago)
Original post by RDKGames
I'm taking an approach with a diagram for help and so I can see that for the open ball Bϵ(c)⊆Rn\ΩB_{\epsilon}(c) \subseteq \mathbb{R}^n \backslash \Omega to not include any element of Ω\Omega, it must be true that my choice for epsilon must instead be defined as ϵ=min⁡0≤i≤k∣xi−c∣\displaystyle \epsilon = \min_{0\leq i \leq k}\vert x_i-c \vert ??


Yep. Though of course epsilon doesn't have to equal that, it can be less than that.
Original post by ghostwalker
Yep. Though of course epsilon doesn't have to equal that, it can be less than that.


Ah cool, thanks.

Now to show that
Unparseable latex formula:

\O

is clopen.

To show that it is closed, I can say that since the set
Unparseable latex formula:

\Omega \backslash \O = \Omega \subseteq \mathbb{R}^n

is open then it's complement,
Unparseable latex formula:

\O

, must be closed.

Then to show it is open, consider A⊂ΩA \subset \Omega that is finite and closed, so
Unparseable latex formula:

A\backslash \O

is closed and the complement,
Unparseable latex formula:

\O

, is open.

Does that seem right?? I've seen different approaches to this (shorter ones too) but very few made sense to me in a simple way to understand.
(edited 7 years ago)
Original post by RDKGames
Ah cool, thanks.

Now to show that
Unparseable latex formula:

\O

is clopen.

To show that it is closed, I can say that since the set
Unparseable latex formula:

\Omega \backslash \O = \Omega \subseteq \mathbb{R}^n

is open then it's complement,
Unparseable latex formula:

\O

, must be closed.


Well we know omega is closed, and not open for a finite set.


Then to show it is open, consider A⊂ΩA \subset \Omega that is finite and closed, so
Unparseable latex formula:

A\backslash \O

is closed and the complement,
Unparseable latex formula:

\O

, is open.

Does that seem right?? I've seen different approaches to this (shorter ones too) but very few made sense to me in a simple way to understand.


My interpretation of the question would be that you want to show ∅\emptyset is clopen (horrible word) in Rn\mathbb{R}^n, and Ω\Omega doesn't come into it.
Original post by ghostwalker
Well we know omega is closed, and not open for a finite set.



My interpretation of the question would be that you want to show ∅\emptyset is clopen (horrible word) in Rn\mathbb{R}^n, and Ω\Omega doesn't come into it.


I'm getting confused now...

How can I start this off? :confused:

EDIT: I want to say that since Rn\mathbb{R}^n is both open and closed then it's complement ∅\emptyset must be closed and open respectively, but I'm not sure if this is sufficient.
(edited 7 years ago)
Original post by RDKGames
I'm getting confused now...

How can I start this off? :confused:

EDIT: I want to say that since Rn\mathbb{R}^n is both open and closed then it's complement ∅\emptyset must be closed and open respectively, but I'm not sure if this is sufficient.


If you know that to be the case, i.e have proved it previously in lectures/exercises, then I would consider it sufficient. Might be worth making sure you're familiar with those proofs.
Original post by RDKGames
I'm getting confused now...

How can I start this off? :confused:

EDIT: I want to say that since Rn\mathbb{R}^n is both open and closed then it's complement ∅\emptyset must be closed and open respectively, but I'm not sure if this is sufficient.


You want to show ∅\emptyset is both a closed set and an open set in Rn\mathbb{R}^n.

Well let's first show it is open. How do you prove a set is open?

Spoiler

Spoiler



To prove it is closed you want to show it's complement is open. But its complement in Rn\mathbb{R}^n is Rn\mathbb{R}^n. So it remains to prove Rn\mathbb{R}^n is open. Take a point in Rn\mathbb{R}^n can we find a ball of positive radius contained in Rn\mathbb{R}^n?
(edited 7 years ago)
Original post by ghostwalker
If you know that to be the case, i.e have proved it previously in lectures/exercises, then I would consider it sufficient. Might be worth making sure you're familiar with those proofs.


Okay it took me a bit of digging but we indeed prove at one point that Rn\mathbb{R}^n is open so I'm happy with this then.

Capture.PNG

For this one, since p∈Rnp \in \mathbb{R}^n is a limit point of Ω\Omega, then Bϵ(p)∩ΩB_{\epsilon}(p)\cap \Omega is neither ∅\emptyset nor equal to {p}\{ p \} for ϵ>0\epsilon > 0. But what can I do with this to start off the proof?

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