Proving the Open Sets Theorem for Metric Spaces

[Buri]

I'm reading Analysis on Manifolds by Munkres and in the section Review of Topology Munkres states the following theorem without proof:

Let X be a metric space; let Y be a subspace. A subset A of Y is open in Y if and only if it has the form A = U ∩ Y where U is open in X.

All he has defined is a metric space, subspace, B(y;ε) = {x|d(x,y) < ε} which is the ε neighborhood of y. And he defined open sets to be: A subset U of X is said to be open in X if for each y ∈ U there is a corresponding ε > 0 such that B(y;ε) is contained in U.

I have never taken Topology or even read about it, so I wrote a proof for it which I'm not sure is correct. Here it is:

Since A is open then for x0 ∈ A there exists an ε > 0 such that B(x0;ε) is in A. Further, B(x0;ε) is open (I've proved this already). Now taking the union of all such ε neighborhoods of x's in A also produces and open set (I've also proved this). Therefore, letting U = U B(x; ε) proves this direction as clearly, A = U B(x;ε) ∩ Y.

(⇐)

A = U ∩ Y

This means that any x ∈ A is also in U and Y. Therefore, since x ∈ U, which is open, there exists B(x; ε) in U. But I want the open ball to be in A so letting ε' = ε and then B'(x;ε') = B(x;ε)∩Y. However since B(x;ε) is in U, then B'(x;ε') is in U∩Y which is A.

Am I on the right track? Btw, what if I have a metric X, and a closed set Y in X and then I choose U to be open and not entirely in Y but entirely in X. I define A to be U∩Y, but then A isn't open since it contains part of the boundary of Y. What am I not understanding? And if X is a metric space and Y a subspace, if A is open in Y, does it necessarily imply that A is open in X also?

I'd appreciate the help!

 

[Martin Rattigan]

I'm reading Analysis on Manifolds by Munkres and in the section Review of Topology Munkres states the following theorem without proof:

Let X be a metric space; let Y be a subspace. A subset A of Y is open in Y if and only if it has the form A = U ∩ Y where U is open in X.

It may be intended as a definition. Given a subset Y of a metric space X there are two routes to defining a topology on it.

(i) metric space X -> topological space X -> topological subspace Y.
(ii) metric space X -> metric subspace Y -> topological subspace Y.

In (i) the open sets for the topological subspace Y are defined as the sets U ∩ Y where U is open in the topological space X.

The final topology on Y is the same via either route which is what you're proving. The proof looks OK (but would be clearer if you made a separate notation such as B_Y(y;ε)=B(y;ε)∩Y for the open spheres in Y).

The open spheres in the metric space X are not necessarily open spheres in the metric subspace Y nor vice versa. Also if B(x;ε) means the open sphere in X with centre x and radius ε, then B(y;ε)∩Y is an open sphere in Y if y∈Y, but B(x;ε)∩Y is not necessarily an open sphere in Y.

 

[Martin Rattigan]

I just read your penultimate paragraph more carefully. What you seem to be misunderstanding is that open sets in the topological space X are not necessarily open sets in the topological space Y nor vice versa. The example you give is open in Y but not in X. (Which also answers your last question).

 

[Buri]

Thanks for your help. I was reading up on topologies on a set X a bit and I feel like I understand how it is possible for something to be open in Y but not in X (if Y is a subspace of X). So open sets are relative to their universe so to speak. However, it seems like the balls B(y;ε) = {x|d(x,y) < ε} are not really dependent on the set they're in. As the proof that B(y;ε) is open (Munkres asks the reader to prove it) follows from the triangle inequality which is part of the definition of a metric and so, they're always open. So why is it that B(y;ε) seems to be special? Is there a reason why we prove sets are open by using these open balls?

Sorry if these are just dumb questions...

 

[Martin Rattigan]

Consider \mathbb{R} as a metric space (with d_{\mathbb{R}}(x,y)=|x-y| for x,y\in\mathbb{R}). Then \mathbb{Q} is a metric subspace (with metric d_{\mathbb{Q}}(p,q)=d_{\mathbb{R}}|_{\mathbb{Q}\times\mathbb{Q}}(p,q)=|p-q|).

But \frac{\sqrt{2}}{2}\in B_\mathbb{R}(0;1) and \frac{\sqrt{2}}{2}\notin \mathbb{Q}, so B_\mathbb{R}(0;1) is not an open ball in \mathbb{Q}.

Similarly B_\mathbb{Q}(0;1) can't be B_\mathbb{R}(x,r) for any x\in \mathbb{R} and r\in \mathbb{R}^+, because it would have to be
B_\mathbb{R}(\frac{\text{sup }B_\mathbb{Q}(0;1)+\text{inf }B_\mathbb{Q}(0;1)}{2};\frac{\text{sup }B_\mathbb{Q}(0;1)-\text{inf }B_\mathbb{Q}(0;1)}{2})=B_\mathbb{R}(0;1)

but \frac{\sqrt{2}}{2}\in B_\mathbb{R}(0;1) and \frac{\sqrt{2}}{2}\notin B_\mathbb{Q}(0;1).

Further B_\mathbb{R}(\sqrt{2};1)\cap\mathbb{Q} is also not an open ball in \mathbb{Q}.

We don't so much prove sets are open using open balls as define open sets in metric spaces using open balls, though of course to prove any particular set in a metric space is open we have to show the definition is satisfied. Open balls don't exist in a topological space where the topology is not derived from a metric.

 

[Buri]

Open balls don't exist in a topological space where the topology is not derived from a metric.

I don't understand what you mean. Could you please explain?

Just another question, I was trying to understand what a topology on a set X is and if I take X = R and have \tau contains all subsets of X then this would be a topology for X, right? Topology on Munkres defines an open set in the following way, a subset U of X with topology \tau is open in X if U is in \tau. So with my example above, then sets which we normally call closed (like [a,b]) would then be called open? This is all because when we normally say (a,b) is open and [a,b] is closed we're using the \tau which only has open sets in the "normal" sense?

 

[Martin Rattigan]

You pretty much explained it yourself in the second paragraph.

If S is a set of cabbages then \mathcal{P}(S)=\{O:O\subset S\} is a topology on S. It is unnecessary to define a distance function d_S(b,c) between cabbages b,c\in S in order to define the topology \mathcal{P}(S). If no such distance function is defined then the normal definition of B_S(c;r) as \{b\in S:d_S(b,c)&lt;r\} becomes meaningless.

If S is any set and B\subset\mathcal{P}(S), then B can be used to generate a topology on S in the same way that the set of open balls generates a topology in a metric space. The open balls are meaningful only in metric spaces.

 

[Buri]

Ahh I get it now. Thanks a lot for your help. I really appreciate it :)