6
$\begingroup$

Is there a topology $\tau$ on $\omega$ such that $(\omega,\tau)$ is Hausdorff and path-connected?

$\endgroup$
20
$\begingroup$

No, a path-connected Hausdorff space is arc-connected, whence it would be of (at least) continuum cardinality provided it has more than one point. This follows from a more general (and deep) result that a Peano space (a compact, connected, locally connected, and metrizable space) is arc-connected if it is path-connected, together with the observation that a Hausdorff space that is the continuous image of the unit interval is a Peano space. See this section of the nLab, and references therein.

$\endgroup$
  • 1
    $\begingroup$ Shouldn't you say instead that it has cardinality at least continuum? It could be far larger, such as with the cone over an enormous discrete space. $\endgroup$ – Joel David Hamkins Aug 11 '15 at 14:14
  • $\begingroup$ @JoelDavidHamkins Oh, of course -- I'll edit in. Thanks! $\endgroup$ – Todd Trimble Aug 11 '15 at 14:28
  • 1
    $\begingroup$ And I guess we have to add the qualifier, "with at least two points," as well, since the one-point space is a path-connected Hausdorff space, but doesn't have size at least continuum. $\endgroup$ – Joel David Hamkins Aug 11 '15 at 15:50
  • $\begingroup$ @JoelDavidHamkins Fixed -- thanks again. $\endgroup$ – Todd Trimble Aug 11 '15 at 17:07
28
$\begingroup$

Todd has already answered the question, but let me give an alternative argument.

Theorem. Every compact Hausdorff space of size less than the continuum is totally disconnected.

Proof. Suppose $a\neq b$ in a compact Hausdorff space $X$ of size less than the continuum. Since every compact Hausdorff space is normal, there is by Urysohn's lemma a continuous function $f:X\to\mathbb{R}$ such that $f(a)=0$ and $f(b)=1$. Since $X$ is smaller than the continuum, there is some real number $z$ strictly between $0$ and $1$ with $z\notin\text{ran}(f)$. So $\text{ran}(f)$ is disconnected and there can be no connected subspace of $X$ containing both $a$ and $b$. So $X$ is totally disconnected. QED

In particular, any such space with at least two points is not connected.

Corollary. There are no nontrivial paths in any Hausdorff space of size less than the continuum.

Proof. The image of a nontrivial path in such a space would be a compact Hausdorff connected space of size less than the continuum, with at least two points, contrary to the theorem. QED

$\endgroup$
  • 2
    $\begingroup$ Very nice! ${}$ $\endgroup$ – Todd Trimble Aug 11 '15 at 15:43
  • 2
    $\begingroup$ Right, Joel - Urysohn's function. Thank you Todd for pointing me to Joel's answer. (Frankly, the q. posed above should never be considered "research"). $\endgroup$ – Włodzimierz Holsztyński Aug 24 '15 at 19:51
13
$\begingroup$

I realise this is quite an old question, but here is an alternative to the already existing elegant answers, which shows that Hausdorff can be replaced by $T_1$:

Proposition. No countable $T_1$ space containing at least two points is the continuous image of an interval.

Proof. Suppose $X$ is a $T_1$ topological space (so points are closed), $I:=[0,1]$, and $f\colon I\to X$ is continuous. Then $$ \{f^{-1}(x)\colon x\in X\}$$ is a partition of $I$ into closed, hence compact, subsets. But $I$ (more generally, any non-trivial continuum) is not the countable disjoint union of compact proper subsets. So $X$ is uncountable. $\square$

Observe that the proof, in contrast to Joel's, does not show that the cardinality of $X$ is that of the continuum. On the other hand, there are non-Hausdorff compact $T_1$ spaces, so his proof does not apply to $T_1$ spaces. Likewise for Todd's argument, as there are $T_1$ spaces that are path-connected, but not arcwise connected.

On the other hand, Sierpiński space (consisting of two points, only one of which is closed) is both $T_0$ and path-connected. So $T_1$ cannot be replaced by $T_0$.

EDIT. It seems this argument is also here: http://topospaces.subwiki.org/wiki/Path-connected_and_T1_with_at_least_two_points_implies_uncountable

$\endgroup$
  • $\begingroup$ Very nice argument, thanks for adding it! $\endgroup$ – Dominic van der Zypen Aug 23 '15 at 8:58

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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