Proof of topological space

Question

My solution:

1) $ cl(A) \subseteq cl(cl(A))$

let $x \in cl(cl(A))$ by definition $U_x \cap cl(A) \neq \varnothing $ $ \forall U_x \in Tau$ such that $x\in U_x $ .Thus $x \in cl(cl(A))$ $cl(A) \subseteq cl(cl(A))$ Is it true? And how prove other side.

Answer 1

To prove $\operatorname{cl}(\operatorname{cl}(A)) \subseteq \operatorname{cl}(A)$ directly using your definition:

Suppose $x \in \operatorname{cl}(\operatorname{cl}(A))$, and suppose we have some open neighborhood $U_x$ of $x$. Then, by the definition, $\operatorname{cl}(A) \cap U_x \ne \emptyset$. Choose some $y \in \operatorname{cl}(A) \cap U_x$. Then, since $y \in U_x$ and $U_x$ is open, we see that $U_x$ is also an open neighborhood of $y$; and since $y \in \operatorname{cl}(A)$, applying the definition again, we get that $A \cap U_x \ne \emptyset$. In conclusion, we have shown that for any open neighborhood $U_x$ of $x$, then $A \cap U_x \ne \emptyset$, which by definition means that $x \in \operatorname{cl}(A)$.

Answer 2

In your proof I don't see why $\mathrm{cl}(A)\subseteq \mathrm{cl}\mathrm{cl}(A)$ follows. You need to start with $x\in \mathrm{cl}(A)$. Let me give you some hints.

First, for every subset $A\in X$, $A\subseteq \mathrm{cl}(A)$ is always true. For every $x\in A$, observe that every open neighborhood $U$ of $x$ will intersect with $A$.

To prove $\mathrm{cl}(A) \supseteq \mathrm{cl}\mathrm{cl}(A)$, prove that $$ \mathrm{cl}(A)= \bigcap_{C\textrm{ closed}\\C\supseteq A}C $$ $\mathrm{cl}(A)$ is a closed set because it is an intersection of closed sets, and since $\mathrm{cl}\mathrm{cl}(A)$ is an intersection of closed sets containing $\mathrm{cl}(A)$, $\mathrm{cl}(A)$ must contain $\mathrm{cl}\mathrm{cl}(A)$.

Answer 3

$cl(A)$ is the intersections of all closed sets which contains $A$ and so $cl(A)$ is closed and obviously $A\subset cl(A)$ $\ldots (1)$.

And if $A$ is closed, since one of the closed sets which contains $A$ is also $A$, we have $cl(A)\subset A$. By (1) we can say $cl(A)= A$$\ldots (2)$.

By using (1), $cl(A)$ is closed, then by using (2) $cl(cl(A))=cl(A)$