A is a subset of a topological space X

• Aug 27th 2011, 05:24 PM
dwsmith
A is a subset of a topological space X
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.
• Aug 27th 2011, 06:38 PM
hatsoff
Re: A is a subset of a topological space X
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.
• Aug 27th 2011, 08:07 PM
dwsmith
Re: A is a subset of a topological space X
Quote:

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?
• Aug 27th 2011, 08:18 PM
hatsoff
Re: A is a subset of a topological space X
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.
• Aug 27th 2011, 08:47 PM
dwsmith
Re: A is a subset of a topological space X
Quote:

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?
• Aug 28th 2011, 03:32 AM
hatsoff
Re: A is a subset of a topological space X
Where did you get the idea that $A\subset U_x$ for each $x\in A$ ?
• Aug 28th 2011, 11:16 AM
dwsmith
Re: A is a subset of a topological space X
Quote:

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.
• Aug 28th 2011, 11:28 AM
Plato
Re: A is a subset of a topological space X
Quote:

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.
• Aug 28th 2011, 01:40 PM
HallsofIvy
Re: A is a subset of a topological space X
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.