Let $X = \mathbb{R}$ with standard topology, $A \subseteq R$ and define an equivalence relation $\mathcal{R}_A$ on $X$ with classes $[x]=\{x\}$ if $x \notin A$ and $[x]=A$ if $x \in A$. Is there a countable subset $A \subseteq \mathbb{R}$ such that the quotient topological space $X/\mathcal{R}_A$ is compact?

Here is my attempt:

No. Suppose $A \subseteq \mathbb{R}$ is countable, then we enumerate the elements as $A=\{a_1,a_2,...\}$ and for $\epsilon > 0$, let

$$\mathcal{A}:=\{(a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i}) \mid i \geq1\}$$

be an open cover for $A$. If we let $p:X \rightarrow X_A$ be the quotient map then

$$X_A \setminus \{p\Big((a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i})\Big) \mid i \geq1\}$$

is non-empty because $\mathcal{A}$ does not cover $\mathbb{R}$ by considering the sum of measures

$$\Big| \bigcup (a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i}) \Big| \leq \sum \Big|(a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i})\Big|< \infty.$$

But we have homeomorphism

$$X_A \setminus \{p\Big((a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i})\Big) \mid i \geq1\} \cong \mathbb{R}\setminus \bigcup(a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i})$$

and since the set on the RHS is not bounded, it is not compact, so we can find an open cover for RHS (hence LHS) without a finite subcover. Then the open cover for LHS together with $\{p\Big((a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i})\Big) \mid i \geq1\}$ is an open cover for $X_A$ which has no finite subcover. $\square$

  • $\begingroup$ For you, a compact space is always Hausdorff? $\endgroup$ – José Carlos Santos Apr 4 '18 at 12:26
  • $\begingroup$ No. These two definitions are independent of each other. $\endgroup$ – Bernoulli Apr 4 '18 at 12:26
  • $\begingroup$ The complement of $A$ is uncountable. Take a discrete infinite countable subset $B$ of the complement of $A$. Take non-intersecting neighborhoods $U_x$ for each element of $B$. Enumerate $A$ and cover its $n$-th element by an open ball $V_n$ of radius $2^{-n}$. Take $W$ to be an open set containing the complement of $\bigcup_{x\in B}U_x$ but not containing any $U_x$. This can be done since $B$ is discrete. Then $S_x=U_x\cup\bigcup_{n=1}^{\infty}V_n$ and $W\bigcup_nV_n$ are the preimages of open sets in the quotient. Those open sets cover the quotient but any finite subset of them doesn't. $\endgroup$ – user547557 Apr 4 '18 at 12:54
  • $\begingroup$ Thank you for your idea. $\endgroup$ – Bernoulli Apr 4 '18 at 17:04

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