Consider the topological space $(\mathbb{R}, \mathcal{T}:=\{U \subseteq \mathbb{R} \mid \forall x \in U: \exists \epsilon > 0: [x, x + \epsilon[ \subseteq U\})$. Is this space separable? Is it first countable? Second countable?

First of all, it looks a lot like this space is metrisable, but I tried some metrics and couldn't find a metric that induces the topology, so I tried to prove it in the hard way.


Let $G$ be a non empty open set. Let $x \in G$ be fixed. Then, there is $\epsilon > 0$ s.t. $[x,x + \epsilon[ \subseteq G$. Invoking the denseness of $\mathbb{Q}$ in $\mathbb{R}$, we can pick $q \in [x, x + \epsilon[ \cap \mathbb{Q}$, and it follows that $\mathbb{Q} \cap G \neq \emptyset$. Hence, $\mathbb{Q}$ is a dense countable set.

Second countable ($\implies$ first countable)

Let $\mathcal{B}:= \{[x,x + \epsilon[ \mid x, \epsilon \in \mathbb{Q}\}$.

First, it seemed reasonable to claim that this is a basis for $\mathcal{T}$.

Indeed, let $x \in B \in \mathcal{T}$. Then, for some $\epsilon > 0$, we have $[x,x+ \epsilon[ \subseteq B$. Choose $y,z \in [x,x + \epsilon[\cap \mathbb{Q}$ with $y < z$. Then $[y,y + (z-y)] \subseteq [x,x + \epsilon[ \subseteq B$. But this doesn't work, as $x \notin [y,y+(z-y)]$ necessarily.

How can I proceed?

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Lower_limit_topology would answer a lot of these questions, I think (though without proofs). Hint for second countable: show that any basis would have to have an element of the form $[x, x + \epsilon_x[$ for each $x \in \mathbb{R}$. $\endgroup$ – Daniel Schepler Apr 25 '18 at 21:42
  • $\begingroup$ What's wrong with your $\mathcal{B}$: for example $[\pi, \pi + 1[$ is open in $\mathcal{T}$ and $\pi \in [\pi, \pi + 1[$, but there is no interval $I$ with rational endpoints such that $\pi \in I \subseteq [\pi, \pi + 1[$. $\endgroup$ – Daniel Schepler Apr 25 '18 at 21:44
  • $\begingroup$ Yes I realised that. Thanks for the link. $\endgroup$ – user370967 Apr 25 '18 at 21:44
  • $\begingroup$ This is also known as the Sorgenfrey line. A separable metrizable space is 2nd-countable. The Sorgenfrey line is separable ($\Bbb Q$ is a dense subset) but not 2nd-countable. $\endgroup$ – DanielWainfleet Apr 26 '18 at 5:10

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