A is a subset of a topological space X

**dwsmith** · August 27th 2011, 05:24 PM

I am having trouble understanding topology. I have read the sections 3 times and hasn't helped.

Could someone explain this question, the methodology to answering it, and how it is done?

Let X be a topological space; let A be a subset of X. Suppose that for each $x\in A$ there is an open set U containing x such that $U\subset A$ . Show A is open in X.

**hatsoff** · August 27th 2011, 06:38 PM

For ease of notation, since each $U$ depends on a corresponding $x$ , let's say that $U=U_x$ . In other words, for each $x\in A$ there is an open set $U_x\subset A$ such that $x\in U_x$ .

With this notation in mind, what you want to do is show that $A=\bigcup\{U_x:x\in A\}$ . Or you can even just state it, since it's straightforward to see.

**dwsmith** · August 27th 2011, 08:07 PM

Originally Posted by hatsoff

For ease of notation, since each $U$ depends on a corresponding $x$ , let's say that $U=U_x$ . In other words, for each $x\in A$ there is an open set $U_x\subset A$ such that $x\in U_x$ .

With this notation in mind, what you want to do is show that $A=\bigcup\{U_x:x\in A\}$ . Or you can even just state it, since it's straightforward to see.

To show $A=\bigcup\{U_x:x\in A\}$ , do we just say there exists a r > 0 s.t. there is a ball of radius r centered at x and A is the union of all the balls?

**hatsoff** · August 27th 2011, 08:18 PM

No. Remember, not every topological space is a metric space. And if the space is not metric, then it won't have any balls.

Think about the construction of $\bigcup\{U_x:x\in A\}$ . We want to show that $\bigcup\{U_x:x\in A\}\subseteq A$ .... AND that $A\subseteq \bigcup\{U_x:x\in A\}$ . Then we will have proved that the two are equal.

How do we show that $\bigcup\{U_x:x\in A\}$ is a subset of $A$ ? Well, remember that each $U_x$ is a subset of $A$ . So the union of all $U_x$ must also be a subset of $A$ .

Now show that $A$ is a subset of $\bigcup\{U_x:x\in A\}$ , and you'll be done.

**dwsmith** · August 27th 2011, 08:47 PM

Originally Posted by hatsoff

No. Remember, not every topological space is a metric space. And if the space is not metric, then it won't have any balls.

Think about the construction of $\bigcup\{U_x:x\in A\}$ . We want to show that $\bigcup\{U_x:x\in A\}\subseteq A$ .... AND that $A\subseteq \bigcup\{U_x:x\in A\}$ . Then we will have proved that the two are equal.

How do we show that $\bigcup\{U_x:x\in A\}$ is a subset of $A$ ? Well, remember that each $U_x$ is a subset of $A$ . So the union of all $U_x$ must also be a subset of $A$ .

Now show that $A$ is a subset of $\bigcup\{U_x:x\in A\}$ , and you'll be done.

Each $x\in A\subset U_x$ . So A is a subset of the big union?

**hatsoff** · August 28th 2011, 03:32 AM

Where did you get the idea that $A\subset U_x$ for each $x\in A$ ?

**dwsmith** · August 28th 2011, 11:16 AM

Originally Posted by hatsoff

Where did you get the idea that $A\subset U_x$ for each $x\in A$ ?

Topology seems to be a struggle for me. That was what I thought. I don't know. I have read the book chapters, I go to class, but I just don't get it.

**Plato** · August 28th 2011, 11:28 AM

Originally Posted by dwsmith

Topology seems to be a struggle for me. That was what I thought. I don't know. I have read the book chapters, I go to class, but I just don't get it.

Topology more than any other undergraduate mathematics is all about definitions. The key then is understanding the ideas behind the definitions.

All this proof is saying is that an open set is the union of other open sets.

**HallsofIvy** · August 28th 2011, 01:40 PM

You were told that "for every x in A there exist an open set, $U_x$ , such that $U_x\subset A$ . You cannot conclude from that " $A\subset U$ "!

What you can conclude is that, for each x in A, x is in $U_x$ and so in the union of all such U sets.

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