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Consider the topological space $\mathbb{R}$ with discrete metric

$$d(x,y) = \begin{cases} 1 & x\neq y\\ 0 & x=y \end{cases}$$

We know that the metric space is first countable. So for each $x\in \mathbb{R}$, there exists a countable neighborhood basis at $x$.

My question is what is that basis?

I think $\{\{x\}\}$ is a local base at each $x$, so there is only one element in this basis? How about if I consider the open ball with $r>1$, then the number of elements in this basis is infinite. It becomes uncountable.

I am confused about how to find the neighborhood basis of $x$.

Martin Sleziak
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sleeve chen
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1 Answers1

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You are right; $\bigl\{\{x\}\bigr\}$ is a a countable (finite, actually) system of neighborhoods of $x$, for each $x\in X$.

Note that, with respect to the discrete topology, $V$ is a neighborhood of $x$ if and only if $x\in V$.

José Carlos Santos
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  • Could you please say more about neighborhood basis for this discrete topology? thanks! – sleeve chen Oct 08 '18 at 07:21
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    That's what my second sentence is about: $\bigl{{x}\bigr}$ is a neighborhood basis for $x$ since each of its elements is a neighborhood of $x$ and since, for each neighborhood $V$ of $x$, $V\supset{x}$. – José Carlos Santos Oct 08 '18 at 07:28