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Given $A \subseteq X$ in the discrete and the trivial topology, find closure of $A$

Note the definition of closure I am using is one in Munkres:

$x \in \overline A \iff \text{ for every open set } U \text{ containing } x, U \cap A \neq \varnothing$

Attempt:

Easy case.

  • Let $(X, \tau_{trivial})$ be the trivial topological space. Let $A \subseteq X$.

  • Then $\overline A = \{ x \in X| \forall U \in \tau_{trivial}, x \in U \implies U \cap A \neq \varnothing\}$

  • Since $\tau_{trivial} = \{\varnothing, X\}$, the only open set that contains $x$ and has non-empty intersection with $A$ is $X$.

  • Hence $\overline A = X$

Case I am having trouble with:

  • Let $(X, \tau_{discrete})$ be the discrete topological space. Let $A \subseteq X$.

  • Then $\overline A = \{ x \in X| \forall U \in \tau_{discrete}, x \in U \implies U \cap A \neq \varnothing\}$

  • Since $\tau_{discrete} = \{A| A \subseteq X\}$

  • ...missing arguments

  • Hence $\overline A = A$

Question:

I realized that given an arbitrary open set $U$ containing $A$, then there could exist an open set $V$ such that $x \in V \subset U, x \notin A$, therefore $\overline A$ necessarily equal to $A$ itself. How do I formally express this in the second proof?

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  • $\begingroup$ In the discrete topology, every set is open. In particular, if $x \in \overline{A} \setminus A$, then $x \in X \setminus A$, which is open, and so $x \in (X \setminus A) \cap A$, a contradiction. Thus, it must be the case that $\overline{A} \setminus A$ is empty. $\endgroup$ – qaphla May 21 '16 at 23:28
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In discrete metric space, every set is open. So for $x$ $\{x\} \cap A \ne \emptyset$ so $x \in A$ so $\overline {A} \subset A$.

And if $x \in A$ and if $x \in U; U$ open then $\{x\} \subset U \cap A$ so $s \in \overline A$. So $A \subset \overline A$. (Trivially true for all topologies.)

So $A = \overline{A}$.

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Here's one way to phrase this: Let $x \in \overline{A}$. Then every open set containing $x$ must intersect $A$ nontrivially. Since $X$ is under the discrete topology, we know that $\{x\} \subseteq X$ is open.

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$\overline{A}$ is the smallest closed subset of $X$ containing $A$. If $X$ has discrete topology, every subset is open hence every subset is closed. So $A$ is closed and so it must be the smallest closed subset containing $A$. We have $\overline{A} = A$.

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