# Thread: Is a compact subset A of a discrete metric space a finite set?

1. ## Is a compact subset A of a discrete metric space a finite set?

Question: Prove that a compact subset A of a discrete metric space is a finite set.

My Attempt: $A$ is a compact subset of a discrete metric space $(S,d)$. Therefore $A$ is both closed and bounded. Since $A$ is bounded there is some Neighborhood $N_r(p)$ such that $A$ is contained in the neighborhood, $A \subseteq N_r(p)$. so, $A \subseteq \{q \in S:d(p,q) ...

And this is where I am getting lost. I know that I have to use the discrete metric $d(p,q) = \left \{ \begin{array}{cc}0, & \mbox{ if } p=q\\1, & \mbox{ if } p!=q\end{array}\right.$ but I'm not sure how because I have never used a discrete metric before.

Any hints, suggestions, or pushes in the right direction would be greatly appreciated.

2. ## Re: Is a compact subset A of a discrete metric space a finite set?

Originally Posted by SleepyGoose
Question: Prove that a compact subset A of a discrete metric space is a finite set.
My Attempt: $A$ is a compact subset of a discrete metric space $(S,d)$. Therefore $A$ is both closed and bounded. Since $A$ is bounded there is some Neighborhood $N_r(p)$ such that $A$ is contained in the neighborhood, $A \subseteq N_r(p)$. so, $A \subseteq \{q \in S:d(p,q) ...
For each $a\in A$ define $O_a=\{x:d(x,a)<0.5\}$.

For each $O_a$ is open set that contains only one point.
But $\{O_a:a\in A\}$ is an open covering of $A$.

Can you finish?

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### a subset A of discrete metric space is compact

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