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Let $X$ be a locally compact topological group. Collapse each connected component to a point. More generally define a equivalence relation on $X$ as, $x \sim y$ iff $x$ and $y$ are contained in a connected subspace. Consider $X/ \sim$, the quotient space of this equivalence relation with quotient topology. Is it totally disconnected?

I have been trying to use the fact that locally compact topological groups are normal, with no progress.

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    $\begingroup$ Are you sure that you want connected components and not path components? $\endgroup$ – Randall Aug 30 '17 at 14:20
  • $\begingroup$ No only connected. $\endgroup$ – Mahbub Alam Aug 30 '17 at 14:24
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The class of $x \in X$ under $\sim$ is exactly the connected component of $x$. This class $[x]$ is connected because if $y \in [x]$ we have $y \sim x$ so $x,y \in C_y$ for some connected set $C_y$. But then $C_y \subset [x]$ by definition and so $[x] = \cup\{C_y: y \in [x]\}$ is connected as a union of connected sets that all intersect in $x$. And also if $C$ is a connected subset containing $x$, all points of $C$ are equivalent to $x$ under $\sim$, as witnessed by $C$ itself, so $C \subseteq [x]$. So $[x]$ is the maximally connected subset containing $x$. Note that all these classes (as components) are closed subsets of $X$.

So it's not "more generally", using $\sim$ and its classes is exactly collapsing all components of $X$ to a point. Then give this quotient $Y = \{[x], x \in X\}$ the quotient topology under $q: X \to Y$ defined by $q(x) = [x]$.

Suppose now that $C \subseteq Y$ is a connected component which has at least two "points", say $[x_1], [x_2] \in C$. Then $q^{-1}[C]$ is closed and connected: the first is clear by continuity of $q$, for the second suppose that $q^{-1}[C]=C_1 \cup C_2$ as a disjoint union of non-empty closed sets. Then $q[C_1]$ is closed, as $q^{-1}[q[C_1]] = C_1$, (and $q$ is quotient) because if $p \in q^{-1}[q[C_1]]$, $[p] \in q[C_1]$, so $[p] = [p']$ for some $p' \in C_1$, and $[p']$ cannot intersect both $C_1$ and $C_2$ by connectedness of $[p']$, so $p \in [p'] = [p] \subseteq C_1$. But then $q[C_1]$ and $q[C_2]$ too are disjoint (a supposed $[z]$ in their intersection, would have $[z]$ disconnected by $C_1, C_2$ in $X$, contradiction) closed (as we saw) subsets of $C$ covering $C$, which cannot be.

This shows that $Y$ is totally disconnected.

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  • $\begingroup$ As C is connected why would I suppose C is disconnected by closed sets? $\endgroup$ – William Elliot Aug 31 '17 at 10:21
  • $\begingroup$ Nice! And yeah I used the term "more generally" vaguely, I was interested in the totally disconnectedness of the quotient. And math.stackexchange.com/users/426203/william-elliot is right, you probably wanted to write $q^{-1} C$ as union of disjoint non-empty closed sets. Anyways, thanks :) $\endgroup$ – Mahbub Alam Sep 6 '17 at 7:48
  • $\begingroup$ I edited the question to have $q^{-1}[C]$ indeed. @WilliamElliot $\endgroup$ – Henno Brandsma Sep 6 '17 at 18:30
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If $X$ is any topological space, then the set $\pi_0(X)$ of connected components of $X$, endowed with the quotient topology given by the natural map $X \to \pi_0(X)$, is a totally disconnected space.

See The Stacks Project, Lemma 5.7.8.

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