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I have seen that $\text{cl}(A\cup B)=\text{cl}(A)\cup \text{cl}(B)$. However I don't see why this is true. I can see the the right to left inclusion, but I can't see the inclusion from left to right. Say that I have an element $x$ contained only in two open sets one that intersects $A\cup B$ only in $A\setminus B$ and and another that intersects only in $B\setminus A$ isn't this a contradiction?

Edit: I have seen the proof but I still can't understand what is wrong with the counterexaple above

Thanks in advance

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  • $\begingroup$ Choose some limit point of LHS and observe that it belong to the RHS. $\endgroup$ – Masacroso Oct 26 '16 at 16:06
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(1) ($\supset$) :: \begin{align*} A \subset A \cup B \implies \text{cl}(A) \subset \text{cl}(A \cup B) \\B \subset A \cup B \implies \text{cl}(B) \subset \text{cl}(A \cup B) \end{align*} therefore yielding that $\text{cl}(A) \cup \text{cl}(B) \subset\text{cl}(A \cup B)$

(2) ($\subset$) ::

The subset $\text{cl}(A) \cup \text{cl}(B)$ is closed and both contains $A$ and $B$, therefore $A \cup B \subset \text{cl}(A) \cup \text{cl}(B)$. $\text{cl}(A \cup B)$ is defined to be smallest closed set which contained $A \cup B$, so that any closed set which contained $A\cup B$ also contains $\text{cl}(A \cup B)$. Therefore $\text{cl}(A \cup B) \subset \text{cl}(A) \cup \text{cl}(B)$.

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  • $\begingroup$ Thanks for answer the proof makes sense but I still can't see what is wrong with my counterexample $\endgroup$ – TheGeometer Oct 26 '16 at 16:26
  • $\begingroup$ But how come $x$ \in cl(A)? I have a neigbourhood of $x$ only intersecting at B\A so this neigbourhood does not intersect A so it shouldn't be a limitpoint of A right? $\endgroup$ – TheGeometer Oct 26 '16 at 16:36
  • $\begingroup$ @TheGeometer Indeed, I've got confused! Define your two open sets to be $U_1$ and $U_2$. Then $U_1 \cap U_2$ is a third open set containing $x$, that neither intersect $A$ or $B$. So $x \not\in \text{cl}(A \cup B)$ ;). So it's not a counterexample. $\endgroup$ – Hermès Oct 26 '16 at 16:44
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    $\begingroup$ Is the proof for the first part valid for an arbitrary collection of sets? $\endgroup$ – H. R. Nov 2 '19 at 12:00
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    $\begingroup$ @H.R. No, consider $\{x \in (0,1)\}$. Then, $(0,1)=\cup_{x} cl\{x\} \subseteq cl(\cup_x \{x\})=[0,1]$. If you post as a separate question; I will poste this as an answer :) $\endgroup$ – AIM_BLB Dec 2 '19 at 15:04
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In your proposed counterexample, you've forgotten that open sets are closed under finite intersection.

So if $x$ has a neighborhood that only meets $A\cup B$ in $B\setminus A$ and a neighborhood that only meets $A\cup B$ in $A\setminus B$, then the intersection of these neighborhoods doesn't meet $A\cup B$ at all.

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Let $x\in Cl(A\cup B)$ then every open set containing $x$ intersects $A\cup B$. Thus for $x \in U_\alpha$ where $U_\alpha$ is open in $X$. If $U_\alpha$ intersects $A$, then $x \in Cl(A)$ else $x \in Cl(B)$ either way $x \in Cl(A)\cup Cl(B)$.

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    $\begingroup$ This is because if $y\in Cl(A)$ iff every open set containing $y$ intersects $A$. $\endgroup$ – Riemann-bitcoin. Oct 26 '16 at 16:10
  • $\begingroup$ I think I have a confusion with the definition. How I know it is that $x$ is a limit point of a subset $S$ of a topological space if every neigbourhood of $x$ intersects with $S$. So in our case if $x$ has a neigbourhood only intersecting $A\B$ and another only intersecting $B\A$ it should be a limit point only of $A\cupB$ and not of A or B.Am I using wrond definition or saying something wrong? $\endgroup$ – TheGeometer Oct 26 '16 at 16:19
  • $\begingroup$ The $Cl(A)$ is the limit points unioned with the set A. So $Cl(A)=A \cup A'$. I believe this is the confusion, though I am slightly confused on what your counterexample is saying. $\endgroup$ – Riemann-bitcoin. Oct 26 '16 at 16:57
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Infinite point sequence from (A ∪ B) contains an infinite subsequence from A or contains an infinite subsequence from B

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  • $\begingroup$ ok but what is wrong wit my counterexample? $\endgroup$ – TheGeometer Oct 26 '16 at 16:04

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