Right now I'm studying countability in topology. My notes provides me a definition of first-countable spaces:

In the following $(X,\tau)$ is a topological space.

Definition If $x \in X$ then $\mathcal{E}(x):=\{E \subset X \mid \exists U \in \tau: x \in U \subset E \}$ is the neighbourhood system of $x$.

Definition: $\mathcal{B}_x \subset \mathcal{E}(x) $ is a local basis with respect to $x$ if for every $U \in \tau$ such that $x\in U$ exists $B\in\mathcal{B}_x$ such that $x\in B \subset U$

Definition: $(X,\tau)$ is first-countable or 1-countable if for each $x \in X$ there is a countable local basis $\mathcal{B}_x$ (so $|\mathcal{B}_x|\leq\aleph_0$).

But I don't know what is a countable local basis. In my notes says "It's clear that $\mathcal{B}_x=\{B(x,r):r\in\mathbb{Q}\}$ is a countable local basis". For me is not so clear, someone can explain what is all stuff?


  • $\begingroup$ What is the adjective you don't understand? Countable? Local? $\endgroup$ – AdLibitum Aug 20 '16 at 19:05
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    $\begingroup$ Buy a countable sets detector. They usually go for somewhere between $\aleph_0$ and $2^{\aleph_0}$ dollars. Sometimes you can find cheaper second hand detectors on eBay... $\endgroup$ – Asaf Karagila Aug 20 '16 at 19:17
  • $\begingroup$ I have a definition of local basis. I just write above. My problem begins with "countable sets". Thank you! $\endgroup$ – Melanctha Aug 20 '16 at 19:25
  • $\begingroup$ @Melanctha I reworked the definitions because they seemed rather imprecise and incomplete to me. If you don't like this roll it back. $\endgroup$ – miracle173 Aug 21 '16 at 10:21
  • $\begingroup$ Thank you @miracle173 .I wrote word by word the definitions that my teacher give to us, and I prefer yours, they are clearer than my teacher's. Maybe his notes are incomplete and create dificulties where there isn't. $\endgroup$ – Melanctha Aug 31 '16 at 7:17

A set $X$ is countable if and only if there is an injection from $X$ to $\mathbb{N}$: that is, iff the elements of $X$ can be "counted." (Note that some texts use "countable" to mean "countable and infinite", so you should check exactly how your book uses the word.)

Now, a local base is a set of open sets such that [property]. Presumably you've already seen the definition of local base; that's why I'm ignoring it here.

A countable local base is . . . just a local base which is countable! So, for example, the set $\{B(x, r): r\in\mathbb{Q}\}$ you mention can clearly be injected into $\mathbb{Q}$ (just map $B(x, r)$ to $r$), and $\mathbb{Q}$ is countable (this is a theorem which your text should prove, or mention, somewhere). Note that a local base being countable or not has nothing to do with it being a local base or not; the two aspects of being a "countable local base" are completely unrelated.

  • $\begingroup$ First of all, thank you. My notes provides me a definition of local base, I'm goingo to put it in the question above here. But what about "detect" countable sets? I just have to invent a bijection between a countable set and my set? $\endgroup$ – Melanctha Aug 20 '16 at 19:16
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    $\begingroup$ @Melanctha Yes. If you have a set $X$, and you want to know if $X$ is countable, you need to do one of two things: (i) Find a bijection between $X$ and a set you already know is countable, or (ii) prove that no such bijection exists (see e.g. the proof that the reals are uncountable). This isn't quite accurate - in principle, you could prove that a bijection between $X$ and some countable set exists, without actually exhibiting such a bijection (e.g. if $X=\{1\}$ if the Riemann hypothesis is true, and $\{1, 2\}$ otherwise) . $\endgroup$ – Noah Schweber Aug 20 '16 at 19:46
  • $\begingroup$ (cont'd) However, that's extremely rare in practice. $\endgroup$ – Noah Schweber Aug 20 '16 at 19:46
  • $\begingroup$ I prefer "countable" to mean "not uncountable". In other words "countable" for "finite or countably infinite". But some authors use "countable" to mean only "countably infinite." $\endgroup$ – DanielWainfleet Aug 21 '16 at 2:44
  • $\begingroup$ @user254665 Note that I cover that in the second sentence of my answer. $\endgroup$ – Noah Schweber Aug 21 '16 at 2:50

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