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This is a follow-up to this question: In a graph, connectedness in graph sense and in topological sense

From Wikipedia

Graphs have path connected subsets, namely those subsets for which every pair of points has a path of edges joining them. But it is not always possible to find a topology on the set of points which induces the same connected sets. The 5-cycle graph (and any n-cycle with n>3 odd) is one such example.

I cannot understand this at all. Surely we are talking about putting a topology on the vertex set of the graph. So if we have a path connected graph, the claim seems to be that it is impossible (for some graphs) to find a topology which is also connected. But what's wrong with the indiscrete topology? I would argue that this is a silly topology-- I'd want to topologically distinguish points. But then, if the vertices of the graph are $\{a,b,c,d,\cdots\}$ then try the topology $\tau=\{ \{a,b,c,d,\cdots\}, \{b,c,d,\cdots\}, \{c,d,\cdots\}, \cdots \}$ seems to distinguish points, and be connected.

So maybe the quote implicitly assumes Hausdorff or something? Then how can it possibly only be true for $n$-cycles with "$n>3$ odd"??

I can only assume I've made a major misreading of this quote...?

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What is wrong with the trivial topology is that it claims that some subsets are connected topologically even though they are not connected in the graph sense. – Henning Makholm Jan 29 '12 at 21:25
Yep, the word "subsets" in the quote is what I missed! Srivatsan's answer also makes this clear. – Matthew Daws Jan 29 '12 at 21:32
up vote 3 down vote accepted

Given an undirected graph $G = (V, E)$, we would like to know whether there exists a topology $\mathscr{T}$ on the vertices, such that for every $S \subseteq V$, the induced subgraph on $S$ is connected if and only if $S$ is connected w.r.t. $\mathscr{T}$ (i.e., the subspace topology on $S$ is connected). And the claim is that no such topology exists for $G = C_{5}$.

The mistake in the OP is that while you are checking the connectedness of the whole set $V$, you are not checking whether the desired condition is met for not every subset $S \subseteq V$.

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Ah...! Yes, now it makes sense! – Matthew Daws Jan 29 '12 at 21:22

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