I know that every non-trivial metric space with more than one point which is connected is uncountable.

However, if we don't demand that the space be a metric space, we can find examples of such odd sets as countable, Hausdorff, and connected spaces.

My question: given a path connected space with more than one point, not necessarily a metric space, can we say anything about the cardinality?

The existence of a countable, connected, and Hausdorff space shows that this is not true if we weaken the statement to just "connected".

What I have tried so far:

According to Wikipedia, if the space is Hausdorff, then path-connectedness implies even arc-connectedness, i.e. we have not only a continuous map from the unit interval to the path, but a homeomorphism, and thus the existence of a bijection between a subset of the space and the unit interval, which implies that the space has at least the cardinality of the continuum.

(Additionally, being arc-connected means that the space must have a one-dimensional topological manifold as a subset, correct? Because of the aforementioned homeomorphism?)


It is unclear to me, however, which suggests that I do not understand the definitions correctly, what happens when the space is not necessarily Hausdorff. If we were talking about a constant path, then the function from the unit interval would necessarily be surjective.

But for any non-constant path, does the function from the unit interval have to be either injective or surjective in order to still be continuous? Does it even have to be bijective, but just not necessarily have a continuous inverse?

Sorry for so many questions; this question kind of developed while I was writing it. If you only answer the first question, I would still be most grateful.

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    $\begingroup$ Take the space $\{a,b\}$ with the open sets $\{a\}$, $\{a,b\}$ and $\emptyset$. Then, map $[0,1)$ to $a$ and $1$ to $b$, and we get that this space is path connected. Though, clearly not Hausdorff. $\endgroup$ – Justin Young Jun 11 '16 at 2:23

Yes, a path-connected space $X$ consisting of more than 1 point has to be uncountable provided it is $T_1$ (you do not need Hausdorff). Otherwise the interval $[0,1]$ can be represented as the disjoint union of (more than one) countably many nonempty closed subsets, which is impossible, see this post.

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  • $\begingroup$ Sorry for being dumb; I don't follow the logic? Where do nonempty closed subsets come into play here? Do you mean that the since the path is compact (being the continuous image of a compact space), if it's in a T1 space then it is also closed? But then for a non-T1 space it no longer has to hold that the path-connected space is uncountable? $\endgroup$ – Chill2Macht Jun 11 '16 at 2:29
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    $\begingroup$ @William: If $f: [0,1]\to X$, define closed subsets in $[0,1]$ to be preimages $f^{-1}(x)$ of points in $f([0,1])$. Is it clear now? $\endgroup$ – Moishe Kohan Jun 11 '16 at 2:32
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    $\begingroup$ @William If $X$ is T1 then points in $X$ are closed subsets of $X$. Then if $f: [0,1] \to X$ is a path connecting two different points, then you can decompose $f$ into $f^{-1}(a)$ (where $a$ varies over all the points in its image); because $f$ is continuous, and points are closed, all of these sets are closed. Thus if $X$ is countable, so is the image of $f$, and we can do the above to decompose $f$ into countably many closed sets, which as pointed out in the answer is impossible. // The comments to your question give a 2-point non-T1 path-connected space. $\endgroup$ – user98602 Jun 11 '16 at 2:32
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    $\begingroup$ @EricWofsey: Somehow I misunderstood the OP and thought that he was assuming that the space in question is Hausdorff. $\endgroup$ – Moishe Kohan Jun 11 '16 at 2:44
  • $\begingroup$ Ohhhh I finally get it now haha thank you all so much! $\endgroup$ – Chill2Macht Jun 11 '16 at 3:06

No, a non-Hausdorff path-connected space with more than one point need not be uncountable. For a trivial example, note that if $X$ has the indiscrete topology, then any map to $X$ is continuous, so trivially any two points can be connected by a path. Less trivially, let $X=\{a,b\}$ with $\{a\}$ open (but $\{b\}$ not open), then for any open set $U\subseteq [0,1]$, the map $f:[0,1]\to X$ sending $U$ to $a$ and $[0,1]\setminus U$ to $b$ is continuous. If $0\in U$ and $1\not\in U$, then $f$ is a path from $a$ to $b$.

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  • $\begingroup$ Ah OK and it isn't even T1, since $b$ isn't closed. It is T0 though, since $a$ and $b$ don't share the same neighborhood systems, correct? So we do need at a minimum T0. $\endgroup$ – Chill2Macht Jun 11 '16 at 3:05
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    $\begingroup$ Yes, my example is $T_0$. But I'm not sure what you mean by "we need $T_0$". $\endgroup$ – Eric Wofsey Jun 11 '16 at 3:49
  • $\begingroup$ actually in hindsight neither do I oops $\endgroup$ – Chill2Macht Jun 11 '16 at 3:52

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