Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 using ZF as my axiom system.

Let $X$ be a metric space and $K\subset Y \subset X$. Suppose $K$ is compact relative to $X$.

Let $\{V_a\mid a\in I\}$ be a family of open sets relative to $Y$ such that $K\subset \bigcup V_a$. Then for every $a\in I$, there exists $G_a$, an open subset of $X$, such that $V_a = Y\cap G_a$.

"I want to choose such $G_a$ for every $a\in I$ without the Axiom of Choice."

I have tried to show whether closure of $X\setminus V_a$ is such $G_a$, but it didn't work well so maybe it's not the way of choosing.

Here's what I have proved.

If $A\subset B$ is open (or closed) relative to $B$, then $B\setminus A$ is closed (or open) relative to $B$.

share|cite|improve this question
Your use of $\LaTeX$ makes your question very difficult to understand. Are you trying to show that $K$ is compact relative to $Y$? Please see the various edits that I and others made to your previous questions, try and learn some more $\LaTeX$. – Asaf Karagila Jul 24 '12 at 11:52
up vote 0 down vote accepted

I suppose that you are trying to show that $K$ is compact relative to $Y$ as well.

You don't need to choose $G_a$ and it might be impossible too, instead consider the open cover of $K$ which consists of all such sets, namely:

$$\{U\subseteq X\mid U\text{ is open, and }\exists a\in I: U\cap Y=V_a\}$$

This is an open cover of $K$ in $X$ and therefore has a finite subcover $U_1,\ldots, U_n$. Now show that $V_i=U_i\cap Y$ is a finite subcover of $\{V_a\mid a\in I\}$.

Generally speaking, the whole point in compactness is that it allows us to avoid the axiom of choice by generating finite sets from which we can choose without the axiom of choice. The trick is often "if you can't choose, take anything that fits, and let compactness choose for you".

share|cite|improve this answer
The last sentence impresses me a lot. Thanks again.. – Katlus Jul 24 '12 at 12:06
@Katlus: This is not a trivial trick, but it's quite common when proving things without the axiom of choice. When you cannot choose and you have a mechanism to make the choices for you, you try to use it in such way. – Asaf Karagila Jul 24 '12 at 12:15

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.