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Let $E, F \subset X$, prove that $ \overline{E \cup F} = \overline {E} \cup \overline{F}$.

For further clarification: I'm referring to $\overline{E}$ as E closure. E' would be the limit point, defined as $E' = \{ ( E \cap N_r (p) ) \backslash \{p\} \neq \emptyset , \ \ \forall R >0 \}$. This is also in a topological space.

I proved it as follows:

$\overline{E \cup F} = \overline{E \cup E' \cup F \cup F'}$ $\\ = \overline{(E \cup E') \cup (F \cup F')}$ $\\ = \overline{(E \cup E')} \cup \overline{(F \cup F')}$ $\\ = \overline{(E)} \cup \overline{F}$

The next one is a little tricky for me. Prove $ \overline{E \cap F} \subset \overline {E} \cap \overline{F}$. Here's my go at it:

Let $x \in \overline{E \cap F}$. Then $x \in \overline{ (E \cup E') \cap (F \cup F)} \rightarrow x \in \overline{E} \cap \overline{F}. $

Feedback would be much appreciated! I've also been trying to come up with examples that would help me visualize these statements a little better.

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  • $\begingroup$ is $\overline{A}$ closure? how do you define it? $\endgroup$ – janmarqz Oct 10 '16 at 0:30
  • $\begingroup$ Sorry, yes! The overline indicates closure. $\endgroup$ – Nikitau Oct 10 '16 at 0:42
  • $\begingroup$ I presume this is meant to be in a topological space? You really should say so if it is. $\endgroup$ – Mitchell Spector Oct 10 '16 at 0:50
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    $\begingroup$ @MitchellSpector Yes I did mean for it to be in a topological space. I will edit my post to reflect this! $\endgroup$ – Nikitau Oct 10 '16 at 0:55
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    $\begingroup$ @janmarqz Yes, thank you! Sorry. I get very clumsy and overlook things with Latex. $\endgroup$ – Nikitau Oct 10 '16 at 0:59
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To prove that $\overline{E \cap F} \subset \overline{E} \cap \overline{F}$:

Let $x \in \overline{E \cap F}$. There exists $(x_n)_n \subset E \cap F$ such that $x_n \underset{n \rightarrow \infty}{\rightarrow} x$.

$(x_n)_n \subset E \cap F$ means that $(x_n)_n \subset E$ and also that $(x_n)_n \subset F$. Can you finish this?

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  • $\begingroup$ Sorry, how do we know that $ \exists (x_n)_n$ such that $x_n \rightarrow_{n\rightarrow \infty} x$? $\endgroup$ – Nikitau Oct 10 '16 at 0:44
  • $\begingroup$ It is equivalent: $x \in \overline{E} \Longleftrightarrow \exists (x_n)_n \subset E$ such that $x_n \underset{n \rightarrow \infty}{\rightarrow} x$. I thought of clarifying this as I didn't know whether or not you knew about this, sorry. $\endgroup$ – Sebastián Oct 10 '16 at 0:48
  • $\begingroup$ You can find another proof of this inclusion in this link: math.stackexchange.com/questions/1149655/…. This uses that the closure of a set is the "smallest" closet set that contains such set. $\endgroup$ – Sebastián Oct 10 '16 at 0:51
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Hint: We have $\overline{E\cup F}=E\cup F\cup (E\cup F)'$, but if you can do $$(E\cup F)'=E'\cup F',$$ then you can get $\overline{E\cup F}=E\cup E'\cup F\cup F'$, so $\overline{E\cup F}=\overline{E}\cup\overline{F}$.

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  • $\begingroup$ Ah, I think I got it. Just to make sure I'm not wrong: I made use of the rule $ A \cap (B \cup C) = A \cap B \cup A \cap C$. Thank you for the hint! $\endgroup$ – Nikitau Oct 10 '16 at 1:25
  • $\begingroup$ for the "union" property, this rule you mention is not employed $\endgroup$ – janmarqz Oct 10 '16 at 1:28

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