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.

  • 3
    $\begingroup$ 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? $\endgroup$ – Daniel Fischer May 27 '14 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? $\endgroup$ – PistolsAtDawn May 27 '14 at 13:10
  • $\begingroup$ It would suffice. But $\mathbb{R}\setminus\mathbb{Q}$ is an even simpler example. $\endgroup$ – Daniel Fischer May 27 '14 at 13:12
  • $\begingroup$ Ok thanks for your suggestions $\endgroup$ – PistolsAtDawn May 27 '14 at 13:16

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? $\endgroup$ – Fardad Pouran Sep 28 '17 at 8:46
  • 1
    $\begingroup$ @FardadPouran a countable base implies separable $\endgroup$ – Henno Brandsma Sep 28 '17 at 12:35

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.