0

if $X$ is a topological space with two equivalence relations $\sim_1,\sim_2$ such that $x\sim_1 y$ implies $x\sim_2 y,$ then if $X/\sim_1$ is compact, then $X/\sim_2$ is compact?

application: From this,$R/Z→R/Q$ is continuous, and sur, image of $R/Z$ (compact) is $R/Q$. We can see $R/Q$ is compact.

  • Consider identity map $\text{Id}\colon X\to X$, and two quotient maps $q_1\colon X\to X/\sim_1$ and $q_2\colon X\to X/\sim_2$. Then, the map $f\colon X/\sim_1\ni [x]1\longmapsto [x]_2\in X/\sim_2$ is well-defined map from hypothesis. Now, $q_2\circ \text{Id}=f\circ q_1$. So, for any $U\subseteq\text{open} X/\sim_2$ we have $(q_2\circ \text{Id})^{-1}(U)\subseteq_\text{open}X$ as $q_2\circ \text{Id}$ is the composition of two continuous map, i.e., $q_2\circ \text{Id}$ is continuous. Hence, $q_1^{-1}\big(f^{-1}(U)\big)=(f\circ q_1)^{-1}(U)= (q_2\circ \text{Id})^{-1}(U)\subseteq_\text{open}X$. – Sumanta Jun 07 '21 at 08:12
  • Now, $q_1$ is quotient map implies $q_1^{-1}(A)$ is open if and only if $A$ is open, in other words, $q_1^{-1}\big(f^{-1}(U)\big)\subseteq_\text{open }X$ implies $f^{-1}(U)\subseteq_\text{open}X/\sim_1$. Therefore, $f$ is continuous. Now, continuous image of a compact space is compact. So, we are done. – Sumanta Jun 07 '21 at 08:12

0 Answers0