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Why is the set of rationals $\mathbb{Q}$, as a subspace of $\mathbb{R}$ with the usual (Euclidean) topology, is not locally compact nor locally connected not locally path connected ??

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closed as off-topic by YuiTo Cheng, Lee David Chung Lin, callculus, verret, The Count Jul 31 at 0:48

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    $\begingroup$ What are your thoughts about the problem? For instance, what are the connected subsets of $\mathbb{Q}$? $\endgroup$ – egreg Jan 24 '14 at 17:39
  • $\begingroup$ All compact subsets of $\mathbb{Q}$ have an empty interior, so none of them are neighborhoods. Hence it follows that no neighborhood of a point in $\mathbb{Q}$ is compact. $\endgroup$ – Tim Seguine Jan 24 '14 at 17:43
  • $\begingroup$ very good Idea yes since Q is dense in (R with usual topology ) i.e cl(Q)=R then there is no subset of Q play as nbd for any point of it hence it's can't be locally compact $\endgroup$ – Taha Topology Jan 24 '14 at 18:17
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    $\begingroup$ I think you misunderstood. It's not that there are no neighborhoods, it's just that none of them are compact. $\endgroup$ – MPW Jan 24 '14 at 18:33
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    $\begingroup$ @TahaGumaelturki $\{1/n : n = 1, 2, 3, ...\} \cup \{0\}$ is compact and infinite. $\endgroup$ – user61527 Jan 24 '14 at 20:03
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for every $a,b \in \mathbb{R}, \space a \neq b$, the open subset $(a,b)_\mathbb{Q} \equiv (a,b) \cap \mathbb{Q}$ can be partitioned given some irrational $c$ for which $a<c<b$ as : $$(a,b)_\mathbb{Q}=(a,c)_\mathbb{Q} \cup (c,b)_\mathbb{Q}.$$ So, there are no connected sets in $\mathbb{Q}$, and therefore it is not locally/connected.

As for local/compactness:
Every open set of the above form can be partitioned to infinite open set elements (Serving as an infinite cover with no finite subcover), and therefore every (non empty) subset of $\mathbb{Q}$ containing an open set is not compact. Given that all neighborhoods contain an open set, it follows that all neighborhoods contained in $\mathbb{Q}$ are not compact. The conclusion is that $\mathbb{Q}$ is not locally/compact.

Also, refer to Arthur Fischer's answer.

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    $\begingroup$ (a,b)<>(a,c)U(c,b) please take care if c belong to (a,b) then (a,b)=(a,c] U(c,b) or (a,b)=(a,c)U[c,b) @Dror $\endgroup$ – Taha Topology Jan 24 '14 at 19:06
  • $\begingroup$ @TahaGumaelturki I didn't understand your message $\endgroup$ – user76568 Jan 24 '14 at 19:07
  • $\begingroup$ It is even totally separated, meaning that for any two points $a,b\in\Bbb Q$ there is a partition of $\Bbb Q$ into two separated subsets. The idea is of course the same, using the irrational $c$ between $a$ and $b$. $\endgroup$ – Stefan Hamcke Jan 24 '14 at 20:59
  • $\begingroup$ @Dror You should say “can be partitioned given some irrational $c$ such that $a<c<b$ as”. $\endgroup$ – egreg Jan 24 '14 at 22:46
  • $\begingroup$ @egreg yes you are right. I thought of that error myself actually in bed, as I closed my eyes to sleep :) $\endgroup$ – user76568 Jan 24 '14 at 23:58
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To show that $\mathbb{Q}$ is not locally-compact, you could make use of the statement from this question:

A subset $M$ of a locally-compact (Hausdorff) space $X$ is itself locally compact iff it is of the form $U \cap F$ for some open $U \subseteq X$ and some closed $F \subseteq X$.1

Now suppose that $U \subseteq \mathbb{R}$ is open and $F \subseteq \mathbb{R}$ is closed such that $\mathbb{Q} = U \cap F$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ and $\mathbb{Q} \subseteq F$, it follows that $F = \mathbb{R}$. But then $U = \mathbb{Q}$, and we know that $\mathbb{Q}$ is not open in $\mathbb{R}$! Therefore $\mathbb{Q}$ is not locally-compact.

(I can't think of a proof of the non-local-connectedness of $\mathbb{Q}$ which substantially differs from Dror's. I will point out that Dror's answer shows that $\mathbb{Q}$ is totally disconnected: the only connected subsets are the singletons. And no non-discrete totally disconnected space is locally-connected.)


1Actually, you don't need to assume that the original space is locally-compact; Hausdorff suffices.

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Another argument for the local compactness : it's not locally compact, because it's not a Baire space, and all locally compact Hausdorff spaces are Baire spaces.

The components of $\mathbb{Q}$ are singletons (because between every two rationals lies an irrational that separates them, essentially), which kills connectedness, and every open subset of $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}$, and so this also holds for all open subsets. So the space is not locally connected (which is equivalent to every component of an open subset is open), because $\mathbb{Q}$ has no isolated points.

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