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If X is a set and T is the collection of all the subsets of X, then all required of a topological space are automatically satisfied. This topology is called the discrete topology. Let X be the real line R. All open intervals (a, b) and their unions define a topology called the usual topology.

Show that, if we allow infinite intersection in the usual topology in R reduces to the discrete topology.

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    $\begingroup$ Hint: For a fixed point $x\in \mathbb{R}$, we have that $\{x\}=\underset{n\in\mathbb{N}}{\bigcap}(x-\frac{1}{n},x+\frac{1}{n})$ is open. $\endgroup$ – ervx Mar 31 '16 at 22:02
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It suffices to allow countable intersections.

Given $a\in \mathbb R$ notice $\{a\}=\bigcap\limits_{n=1}^{\infty}(a-\frac{1}{n},a+\frac{1}{n})$. Being a countable intersection of open sets, $\{a\}$ is open.

Now take $A\subseteq \mathbb R$ and notice $A=\bigcup\limits_{a\in A}\{a\}$. Being a union of open sets, $A$ is open. So every subset of $\mathbb R$ is open, therefore the topology is discrete.


What do we need to get this proof to work? we need to prove the singleton set $\{a\}$ is open. To do this we need to be able to find, for each other element $b$, an open set $U$ so that $a\in U$ and $b\not\in U$. So this works for any $T_1$ space. Note that we need arbitrary intersections in some cases.

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