# Baby rudin 2.34

Baby Rudin 2.34: Prove that a compact subset of a metric space is closed.

I think I have an alternative solution for Rudin 2.34. So can you check whether my steps are correct? Let $$p$$ be a limit point of a set $$K$$. Take any neighborhood $$V_s(p)$$ in metric space $$X$$. This neighborhood must contain some element $$q_1$$ belonging $$K$$. Take a neighborhood $$V_{r_1}(q_1)$$ where $$r_1=d(p,q_1)/2$$. Now take neighborhood $$V_{r_1}(p)$$. This contains an element $$q_2$$ belonging $$K$$. Take a neighborhood $$V_{r_2}(q_2)$$ where $$r_2=d(p,q_2)/2$$. And so on. For other elements of $$K$$ cover them with an open ball. So we have constructed an open cover. But this cover cannot have finite subcover; as if we have then our sequence of neighborhoods stops at some point which means $$p$$ is not a limit point. Contradiction.

• You should start with the problem description. Not everybody has that book on their desk. – Martin R Jul 9 at 15:07
• Edited. Can see it now. – Wu Lai Jul 9 at 15:19

I agree with Siong Thye Goh that it's not a valid proof, but I'm going to rewrite it to give detail where it goes wrong.

Let $$K$$ be a compact subset of a metric space, and assume for the sake of contradiction that it is not closed. Then there exists a limit point $$p\notin K$$. We will use $$p$$ to construct an open cover, and then we will try to show that this open cover has no finite subcover, for a contradiction.

Now construct a sequence of open sets $$U_i$$ as follows. Take $$q_1$$ any point in $$K$$, and let $$U_1$$ be the open ball $$V_{r_1}(q_1)$$ with $$r_1=d(p,q_1)/2$$. Then $$V_{r_1}(p)$$ is disjoint from $$U_1$$ and contains another point $$q_2\in K$$. Let $$U_2$$ be the neighborhood $$V_{r_2}(q_2)$$ where $$r_2=d(p,q_2)/2$$, and note that $$q_1\notin U_2$$. Iterating, we get a sequence of sets $$U_i$$ and points $$q_i$$ such that $$q_i\notin\bigcup_{j\neq i} U_j$$ for all $$i$$.

(So far this is OK -- by the last sentence, we can't remove any of the $$U_i$$ and still cover $$K\cap(\bigcup_i U_i)$$.)

Now comes the hard step: we need an open cover $$\{V_\alpha\}$$ of $$K\setminus (\bigcup_i U_i)$$ such that for all $$i$$ there exists a point $$x_i\in U_i$$ so that $$x_i\notin \bigcup_\alpha V_\alpha$$ -- otherwise we could start removing some sets $$U_i$$ from the open cover, and possibly arrive at a finite subcover. Just taking $$V_\alpha$$ to be open balls for all $$x_\alpha\in K\setminus (\bigcup_i U_i)$$ as you have above does not guarantee this.

• That is right... Thank you very much for both of you! – Wu Lai Jul 9 at 15:58

It is not a valid proof.

You want to prove that $$p$$, the limit point is an element of $$K$$ starting from the assumption that $$K$$ is compact.

You have constructed an open cover for $$K$$, by the assumption that $$K$$ is compact, we must have a finite subcover.

Let be even more concrete and consider $$K=[0,1]$$ a subset of real number and let $$q=0.5$$. We pick the first neighborhood to have radius $$2$$, then we can have $$q_1=1$$. We could have use compactness to pick this neighborhood to be our finite subcover.