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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.

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

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

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