Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm reading Intro to Topology by Mendelson.

The problem statement is in the title.

My attempt at the proof is as follows:

Let $\lbrace U_\alpha\rbrace_{\alpha\in I}$ be an open covering of $X$ such that for each $\alpha\in I$, $U_\alpha\in\mathfrak{J}$. Since $\{U_\alpha\}_{\alpha\in I}$ is also a covering of $X$ in the topology $\mathfrak{J'}$, there then exists a finite subcovering $\{U_{\alpha_i}\}_{i=1}^n$ of $X$, whereby $U_{\alpha_i}\in\mathfrak{J}$ for each $0\leq i\leq n$. Thus, $(X,\mathfrak{J})$ is compact.

My only concern is the last sentence which says all of the subcovering came from the topology $\mathfrak{J}$.

Thanks for any help or feedback!

share|cite|improve this question
Your proof is correct in all respect. – Shailesh Aug 11 '13 at 9:27
Awesome, thanks for the verification. – Shant Danielian Aug 11 '13 at 9:28
up vote 2 down vote accepted

Your proof is correct.

Sooner or later you will learn that continuous images of compact sets are compact. Then this question can be seen as a special case since $\mathfrak J\subseteq\mathfrak J'$ means that $(X,\mathfrak J)$ is the continuous image of $(X,\mathfrak J')$ under the identity map.

share|cite|improve this answer
Thanks for the feedback. I wasn't 100% sure that I could have done what you mentioned, but you made it clear I can. – Shant Danielian Aug 11 '13 at 22:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.