1
$\begingroup$

I need to determine which of the following are true and prove it... if it is false then I have to give a counterexample.

If $A$ is a subset of a topological space, then $A' \subseteq A$

versus

For any closed subset $A$ of a topological space, $A' \subseteq A$.

I think this theorem is helpful "Let $(X, \mathfrak T)$ be a topological space and let $A \subseteq X$. The set $A$ is closed iff $A' \subseteq A$.

$A'$ represents limit points my defintion for limit points is: Let $(X, \mathfrak T)$ be a topological space and let $A \subseteq X$. A point $x \in X$ is said to be a limit point of $A$ provided that every open set containing $x$ contains a point of $A$ different from $x$.

I think the first statement is true and the second statement is false.

$\endgroup$
  • $\begingroup$ To be clear, for any set $A$, you define $A'$ to be the set of limit points of $A$? $\endgroup$ – MartianInvader Apr 28 '15 at 20:41
  • $\begingroup$ Yes, I should have included that in the definition. $\endgroup$ – user219081 Apr 28 '15 at 20:41
  • $\begingroup$ The second statement is true and the first statement is false. Can you see why? Think of subsets of $\mathbb{R}$. $\endgroup$ – Mnifldz Apr 28 '15 at 20:42
  • $\begingroup$ I was thinking the first was false and trying to use the following example in the usual topology $[0,1)$ because 1 is a limit point but 1 is not in the set $[0,1)$ $\endgroup$ – user219081 Apr 28 '15 at 20:45
  • $\begingroup$ even in $\mathbb R$ with the standard topology, take an open interval $(a,b)$ can you see limit points that are not in this set? $\endgroup$ – sha Apr 28 '15 at 20:48
0
$\begingroup$

Hint: for the second claim (the first one is as in the comments) try to assume that for some topological space $(X,\tau )$ and some closed set $B$ in it, there is a limit point of $B$ such that $x\notin B$ thus $x\in B^c$, but what can you say about $B^c$?

Now look at the definition of a limit point. And you will get a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy