Compact metric spaces can be covered by finitely many $\epsilon$-balls

Question

Let $(X,d)$ be a compact metric space, show:

i) For any $\epsilon>0$ there are $n \in \mathbb N$ and $ x_{1},..., x_{n} \in X $ so that $X=\bigcup_{j=1}^{n}B_{\epsilon}(x_{j})$

My thoughts so far (not much):

Since $X$ is a compact space there exists $\bigcup_{j=1}^{N}U_{\lambda_{j}}$ such that $X \subset \bigcup_{j=1}^{N}U_{\lambda_{j}}$.

A huge help would be to say that $\bigcup_{j=1}^{N}U_{\lambda_{j}} \subset X$, but that does not hold right?

Answer 1

HINT: if $X$ is compact then for any open cover there is a finite subcover. Now just choose the open cover defined by $\mathcal B(\epsilon):=\{\Bbb B(x,\epsilon):x\in X\}$ for any chosen $\epsilon>0$.

Answer 2

Since you don't say what the $U_{\lambda_j}$'s are, nothing much can be said about your attempt.

Just consider the set$$\left\{B_\varepsilon(x)\,\middle|\,x\in X\right\}.$$It's an open cover of $X$ and $X$ is compact. Therefore…