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Give an example of a metric space which does not have a countable basis.

I was thinking of some uncountable set, with a metric which results in an uncountable number of open subsets. Which resulted in this:

$$(\mathbb{R}, d_{\text{discrete}})$$ Where $$d_\text{discrete}(x,y) = \begin{cases} 1 & \text{if } x\not = y\\ 0 & \text{if } x=y \end{cases}$$

If I understand this metric correctly then the induced topology $\tau = \{ U\subseteq \mathbb{R}: U \text{ is }d_\text{discrete}\text{-open}\}$ would consists of all subsets of $\mathbb{R}$. And then this would include an uncountable number of sets.

Intuition tells me this cannot have a countable basis. But how do I prove this? Would the following be valid? It doesn't feel very formal...

Proof?

Say there is a countable basis $\mathscr{B}$. Then $(\forall U \in \tau)(U = \bigcup_i B_i)$ where $B_i\in \mathscr{B}$.

But since $\mathscr{B}$ is countable, then $U$ must be countable which leads to a contradiction. ($U$ can be uncountable, like $U=[0,1]$)

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Your example is fine, but your argument is not: $\Bbb R$ with the usual topology, for instance, is second countable but has uncountably many open sets. Let $\mathscr{B}$ be a base for the discrete topology on $\Bbb R$. For each $x\in\Bbb R$ the set $\{x\}$ is open, so for each $x\in\Bbb R$ there is a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq\{x\}$. Clearly this means that $B_x=\{x\}$, so $\{x\}\in\mathscr{B}$ for each $x\in\Bbb R$.

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  • $\begingroup$ Okay, so by construction you show that $\mathscr{B}$ must contain uncountably many elements? $\endgroup$ – dietervdf May 17 '15 at 22:09
  • $\begingroup$ @dietervdf: Exactly. I’m really proving the general fact if $\mathscr{B}$ is a base for $X$, and $x$ is an isolated point of $X$, then $\{x\}$ must be a member of $\mathscr{B}$. $\endgroup$ – Brian M. Scott May 17 '15 at 22:15

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