Looking for some guidance on two topology questions:

(a) Show that a locally connected space with a countable basis, has at most countably many connected components.

(b) Give an example when X has countable basis but it has uncountable many connected components.

Mainly stuck on (b). I think I understand why this is the case for (a) along the lines that for $d$ in a set $D$ (with a countable basis), given $C_{d}$ a connected component for $d$, the intersection between the set of such countable components, $C$, and $D$ is non-empty i.e. $C \cap D \neq \emptyset$, but this means that, by the countability of $D$'s basis, we cannot have uncountably many connected components (elements in the intersection). For (b) I've been casting about and have seen some references in texts to sets $B_{\rho,\theta}$ which have uncountably many connected components but countably many components that are copies of the Mandelbrot set...but suspect this may be over-complicating.

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    $\begingroup$ Every subspace of a space with a countable basis is also second countable. $\mathbb{R}$ for example has a countable basis. Can you find a subspace of $\mathbb{R}$ with uncountably many connected components? $\endgroup$ Commented May 27, 2014 at 12:18
  • $\begingroup$ Would the Cantor set suffice - each point consists of a single point that is connected (it is totally disconnected because its connected components are single points, but each such point remains a connected component) but there are uncountably many of them? $\endgroup$ Commented May 27, 2014 at 13:10
  • $\begingroup$ It would suffice. But $\mathbb{R}\setminus\mathbb{Q}$ is an even simpler example. $\endgroup$ Commented May 27, 2014 at 13:12
  • $\begingroup$ Ok thanks for your suggestions $\endgroup$ Commented May 27, 2014 at 13:16

2 Answers 2


Try $\mathbb{R}-\mathbb{Q}$


As to (1), let $D$ be a countable dense subset of $X$. In a locally connected space $X$, all components $C_x$ of a point $x \in X$ are open sets, and different components are disjoint.

Every distinct component contains a point of $D$ (as components are open and $D$ is dense), and this defines an injective function from the set of all different components into $D$, so the set of components is at most countable.

The irrationals or the Cantor set show that there are second countable, uncountable spaces, where the set of components is all singletons.

  • $\begingroup$ Why such a $D$ exist? $\endgroup$ Commented Sep 28, 2017 at 8:46
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    $\begingroup$ @FardadPouran a countable base implies separable $\endgroup$ Commented Sep 28, 2017 at 12:35

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