Thread: Prove that the set of limit points is closed

1. Prove that the set of limit points is closed

I have a few questions - they all look very straightforward but I haven't actually got answers to them and it's greatly annoying me...

Let (X,T) be a Hausdorff topological space and let A be a non-empty subset of X.

1) Prove that if an open set U in T has non-empty intersection with $\displaystyle \overline A$, where $\displaystyle \overline A$ is the closure of A, then U has a non-empty intersection with A.

2) Prove that A' is closed in A, where A' is the set of limit points of A. (Note, if we can prove that A ' contains all its limit points then we're done, but just because A ' contains all of the limit points of A it doesn't necessarily mean it contains all the limit points of A'...)

3) Prove $\displaystyle (\overline A)' = A'$

2. 1. This can be proven by rewriting the statement of the theorem as, "If $\displaystyle U$ is an open set of $\displaystyle T$ that does not intersect $\displaystyle A$, then $\displaystyle U$ does not intersect $\displaystyle \overline{A}$," using the definition of $\displaystyle \overline{A}$ in terms of open sets.

3. First, $\displaystyle A'\subseteq(\overline{A})'$, for any open neighborhoods of a point minus that point intersecting $\displaystyle A$ must also intersect $\displaystyle \overline{A}$. Now, all we need to prove is that any element of $\displaystyle (\overline{A})'$ belongs to $\displaystyle A'$. For this, we can use the results of (1).

3. 2.

did you get an answer for number 2?