# Thread: Proof related with compact sets

1. ## Proof related with compact sets

Let (X,p) be a compact metric space. Let C be a collection of open subsets of X such that C covers X, i.e., X= Union of Q where Q belongs to C. Prove that there exists a strictly positive real number r, such that for each point a in X, there is a member Q of the collection C such that an open ball Br(a) of radius r centered at 'a' is contained in Q, i.e. Br(a) is a strict subset of Q. Any ideas how to do this question? Thanks!

2. ## Re: Proof related with compact sets

first

$$a \in X \Rightarrow \exists Q \in C \ni a \in Q$$

or else C wouldn't cover X

since the Q's are open

$$a \in Q \Rightarrow \exists r>0 \ni B_r(a)\in Q$$

that should be pretty much it. I leave it to you to finish.

Note of course that if a is in the boundary of X then it will necessarily lie in a Q that is not entirely contained in X but that doesn't matter here.

3. ## Re: Proof related with compact sets

Hey thanks a lot. What did you mean by "Q that is not entirely contained in X" in the last statement? The question says that the Q is an open subset of X, so it must entirely lie in X, shouldn't it? Also is there anything left to prove at the end where you stopped the proof?

4. ## Re: Proof related with compact sets

Ok, my mistake. The Q's are subsets of X.

Actually in this case the problem statement isn't true.

X is compact so it's closed and thus contains all it's boundary points.

But any points on the boundary cannot be enclosed in a B(r) sphere entirely contained in X or they wouldn't be boundary points. So any B(r) sphere around a boundary point can't be a subset of a Q as Q lies entirely in X.

Maybe I'm missing something. This sort of abstract math isn't my forte.

5. ## Re: Proof related with compact sets

Originally Posted by mynameisanthonyg2013
Let (X,p) be a compact metric space. Let C be a collection of open subsets of X such that C covers X, i.e., X= Union of Q where Q belongs to C. Prove that there exists a strictly positive real number r, such that for each point a in X, there is a member Q of the collection C such that an open ball Br(a) of radius r centered at 'a' is contained in Q, i.e. Br(a) is a strict subset of Q. Any ideas how to do this question? Thanks!
Allow me to restate the question: Suppose that $\mathcal{O}$ is an open cover of $X$. Then $\exists r>0$ such that $\forall x \in X$ the open ball $\frak{B}_r(x)$ is a subset of some member of $\mathcal{O}$.

Suppose that is false. Then $(\forall n\in \mathbb{Z}^+)[\exists x_n\in X]$ such that $\frak{B}_{1/n}(x_n)$ is not a subset of any member of $\mathcal{O}$.

Now if $y\in X$ then $\exists U_y \in \mathcal{O}$ $~\&~\exists r_y>0$ such that $\frak{B}_{r_y}(y)\subset U_y$.

I will leave you to finish.
If $1/n < r_y$ then $x_n\ne y$ WHY?
The infinite set $\{x_n\}$ cannot have a limit point. WHY\$
How does this give a contradiction?

BTW: This is known as the Lebesgue number or Lebesgue net for compact spaces.