Is there a metric characterization of connectedness? I'm looking for something like the following metric characterization of compactness: A metrizable topological space is compact if, and only if, every metric inducing the topology is complete and totally bounded.

So, is there any property $P$ of a metric such that a metrizable topological space is connected if, and only if, every metric inducing the topology has property $P$.

If no such $P$ exists, is there a property $P$ such that every meteric space satisfying it is connected, and a metrizable topological space is connected provided there is some metric inducing the topology and which satisfies $P$.

  • $\begingroup$ I'm not sure I understand what kind of condition you're looking for. You can obviously take $P$ to be "the topology induced by the metric is connected"... What kind of property $P$ would be acceptable or not? $\endgroup$ – Najib Idrissi Nov 1 '14 at 14:30
  • $\begingroup$ @NajibIdrissi I'm looking for a metric property, just like the one for compactness I mention above. $\endgroup$ – Ittay Weiss Nov 1 '14 at 17:52

There is for compact spaces:

A compact metric space $X$ is connected if and only if for all $a,b\in X$ and all $\varepsilon>0$ there exist points $a=p_0, p_1, \ldots, p_k=b$ such that $d(p_i,p_{i+1})<\varepsilon$.

And it's not hard to prove this. It's difficult for me to imagine there's some general characterization given examples of connected spaces like the Knaster–Kuratowski fan.

  • $\begingroup$ Thank you, this is helpful, particularly if compactenss can be avoided. $\endgroup$ – Ittay Weiss Nov 8 '14 at 21:31
  • 1
    $\begingroup$ To complement this answer, the above property is known as uniformly connected, or Cantor connected (even if the wikipedia link does not make this obvious) and you might get some references if you google the above terms. $\endgroup$ – Mirko Nov 15 '14 at 15:39

Let us first consider a different characterization of compactness: a metrizable space is compact iff the range of every metric is compact. Necessity follows from continuity of the metric, and sufficiency can be obtained from the fact that every non-compact metrizable space admits an unbounded metric.

This suggest the analogous statement: a metrizable space is connected iff the range of every metric is connected. This turns out to be true, and even simpler.

Necessity again follows from continuity of the metric. To show sufficiency, let $(X, \rho)$ be a disconnected metric space and let $A, B$ be a separation of $X$. We can define a topologically equivalent metric by $$ \sigma(x, y) = \cases{ \min\{ 1, \rho(x, y) \} & if $x,y \in A$ or $x, y \in B$,\\ 2 & otherwise. } $$ Clearly the range of $\sigma$ is not convex, since it contains 0 and 2, but not 3/2, therefore it is disconnected.

Corollary: a metrizable space is a continuum iff the range of every metric is a closed bounded interval.

P.S.: There is a question you did not ask, but which is suggested by your example: is there a property $P$ of metrics such that the existence of a metric with property $P$ implies connectedness and connectedness implies that all metrics have property $P$. Equivalently: is there a property $P$ such that a metric space is connected iff its metric has property $P$. I don't have an answer to that.

  • $\begingroup$ Thanks for this answer. I'm looking for something that is less dependent on special properties of the reals. $\endgroup$ – Ittay Weiss Nov 8 '14 at 21:33
  • $\begingroup$ Depending on exactly which properties you consider special, that may not be a realistic goal. The fact that nontrivial connected metric spaces exist at all is intimately related to the fact that the codomain of a metric is locally homeomorphic to $\mathbb{R}$. $\endgroup$ – Niels J. Diepeveen Nov 10 '14 at 1:13
  • $\begingroup$ I understand this is somewhat vague. By comparison to the compactness case, the metric characterization (complete and totally bounded) is a more inherently metric property than saying "the image of the metric is complete in $\mathbb R$ with the Euclidean topology". I hope this makes things slightly less vague. $\endgroup$ – Ittay Weiss Nov 10 '14 at 3:24
  • $\begingroup$ @IttayWeiss I still don't see what you don't like about this answer. "Complete and totally bounded" fundamentally uses the structure of $\mathbb{R}$; are you looking for a characterization of connectedness which doesn't look like it uses any properties of $\mathbb{R}$? $\endgroup$ – Paul Siegel Nov 12 '14 at 20:02
  • $\begingroup$ @PaulSiegel of course under the hood things will depend on properties of the reals. "Complete and totally bounded" uses the reals, bot not (directly) their completeness or total boundedness. Not to mention the fact that I'd rather not to use the concept of connectedness (of the reals) for defining connectedness for general metric spaces. $\endgroup$ – Ittay Weiss Nov 12 '14 at 20:25

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.