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For any two sets $A$ and $B$ in a space $X$, $$\text{cl}(A \cup B)=\text{cl}(A) \cup \text{cl}(B)$$ where $\text{cl}(Y)$ denote the closure of $Y$

This question and its answer are already available in this site. Why I am posting here is "to check my proof". Here's My try:

$$x \in \text{cl}(A \cup B) \Leftrightarrow \exists \;r>0 \;\text{such that}\;B_r(x) \cap (A \cup B) \neq \emptyset$$

$$\Leftrightarrow (B_r(x) \cap A)\cup (B_r(x) \cap B) \neq \emptyset$$

$$\Leftrightarrow (B_r(x) \cap A) \neq \emptyset \;\text{or}\;(B_r(x) \cap B) \neq \emptyset$$

$$\Leftrightarrow x \in \text{cl}(A)\; \text{or}\; x \in \text{cl}(B)$$

$$\Leftrightarrow x \in \text{cl}(A) \cup \text{cl}(B)$$

Is this right?

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    $\begingroup$ @Cataline The proof is actually "too" concise as it makes a mistake by losing the $\exists r$ in every one of the statements. $\endgroup$ – 5xum Aug 13 '18 at 11:07
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    $\begingroup$ according to your first $\iff$ every $x$ is an element of $\mathsf{cl}(A\cup B)$ $\endgroup$ – drhab Aug 13 '18 at 11:07
  • $\begingroup$ Oh! replacing $\exists$ by $\forall$ is then true? $\endgroup$ – user444830 Aug 13 '18 at 11:09
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I take issue with your first equivalence. Take the open ball of radius $1$ and center $0$. Then for a point $p$, a ball $B_r(p)$ intersects $B_1(0)$ if you take $r = \mathrm{dist(p,0)}$. Is the closure of $B_1(0)$ the full space $\mathbb R^n$ ?

The equivalence you should have is:

$$ x\in \mathrm{cl}({x}) \Leftrightarrow \forall r > 0: B_r(x) \cap(A \cup B) \ne \varnothing $$

But you have an issue in the sense that for a given $x$, $B_r(x)$ will intersect either with $A$ or $B$ and so far that depends on $r$. But you must have either $A$ or $B$ that intersects with $B_r(x)$ for all $r > 0$, not some of them depending on $r$, so you have additional steps to do.

Note that if there is a radius $r_1$ such that $B_{r_1}(x)$ intersects $A$ but not $B$, for any $0 < r_2 < r_1$, $B_{r_2}(x)$ cannot intersect $B$, else you contradict the conditions on $r_1$. Moreover, for $r_3 > r_1$, $B_{r_3}(x)$ must intersect $A$ since $B_{r_1}(x) \subset B_{r_3}(x)$. Thus, you can see that for a given $x$, if there is an $r_1$ such that $B_{r_1}(x)$ does not intersect $B$, then for all $r>0$, $B_{r}(x)$ intersects $A$.

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  • $\begingroup$ So my proof is correct only when replacing $\exists$ by $\forall$ ? $\endgroup$ – user444830 Aug 13 '18 at 11:15
  • $\begingroup$ No. I edited to add why not. $\endgroup$ – Lærne Aug 13 '18 at 11:17
  • $\begingroup$ Thanks! I understand....(Finally my proof becomes one of the mathematical fallacies !) $\endgroup$ – user444830 Aug 13 '18 at 11:24
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This answer shows that the equality is valid in every topological space.

From $A\subseteq A\cup B$ it follows directly that $\mathsf{cl}(A)\subseteq\mathsf{cl}(A\cup B)$.

Similarly we find $\mathsf{cl}(B)\subseteq\mathsf{cl}(A\cup B)$ and conclude that:

$$\mathsf{cl}(A)\cup\mathsf{cl}(B)\subseteq\mathsf{cl}(A\cup B)$$

If $x\notin\mathsf{cl}(A)\cup\mathsf{cl}(B)$ then open sets $U,V$ exist that contain $x$ as element and have an empty intersection with $A$ and $B$ respectively.

Then $U\cap V$ is an open set that contains $x$ as element and has an empty intersection with $A\cup B$. Proved is then that $x\notin\mathsf{cl}(A)\cup\mathsf{cl}(B)$ implies that $x\notin\mathsf{cl}(A\cup B)$. This justifies the conclusion that also:$$\mathsf{cl}(A\cup B)\subseteq\mathsf{cl}(A)\cup\mathsf{cl}(B)$$and we conclude that:$$\mathsf{cl}(A\cup B)=\mathsf{cl}(A)\cup\mathsf{cl}(B)$$


More shortly the second part follows also from the following reasoning: $\mathsf{cl}(A\cup B)$ is the smallest closed set that contains $A\cup B$ as a subset. The sets $\mathsf{cl}(A)$ and $\mathsf{cl}(B)$ are both closed, and consequently their union is closed. Further this union contains $A\cup B$ as a subset so we are allowed to conclude that $\mathsf{cl}(A\cup B)$ is a subset of this union.

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  • $\begingroup$ Nice answer! thanks! $\endgroup$ – user444830 Aug 13 '18 at 11:31

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