Is discrete topology countable?
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Let $$\left( {X,\tau } \right)$$ be a topological space, then $$X$$ is said to be the first countable space if for every $$x \in X$$ has a countable local base; i.e., if every $$x \in X$$, $${{\rm B}_x}$$ is countable. In other words, a topological space $$\left( {X,\tau } \right)$$ is said to be the first countable space if every point $$x$$ of $$X$$ has a countable neighbohood base. A first countable space is also said to be a space satisfying the first axiom of countability. Example: If $$X$$ is finite, then $$\left( {X,\tau } \right)$$ is first countable space. As $$X$$ is finite, all of its subsets are finite. If $${{\rm B}_x}$$ is a local base of $$x \in X$$, then $${{\rm B}_x}$$ is also finite. So, $$\left( {X,\tau } \right)$$ is the first countable space. Example: Base at $$a$$ $$ = {{\rm B}_a} = \left\{ a \right\}$$ Base at $$b$$ $$ = {{\rm B}_b} = \left\{ b \right\}$$ Base at $$c$$ $$ = {{\rm B}_c} = \left\{ {c,d} \right\}$$ Base at $$d$$ $$ = {{\rm B}_d} = \left\{ {c,d} \right\}$$ Here each local base is countable, so $$\left( {X,\tau } \right)$$ is the first countable space. Example: If $$X$$ is either countable or uncountable and $$P\left( X \right)$$ is a discrete topology on $$X$$, then $$\left( {X,\tau } \right)$$ is always the first countable space, because for each $$x \in X$$, $${{\rm B}_x}$$ (singleton) is the local base and so the local base is finite. $${{\rm B}_x} = \left\{ {\left\{ x \right\}} \right\}$$, $$x \in {{\rm B}_x} \subseteq U$$ and $${{\rm B}_x}$$ is countable (finite).
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I saw a proof on proofwiki but couldn't understand it. I only know the bare basics of topology (closed, open, limit points, neighbourhoods). Is there a simple proof for this? $\endgroup$ 2
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I know that there are countable topological spaces with the discrete topology, which are not second countable. But I cannot find an argument, why uncountable spaces with the discrete topology should be second countable.
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