# Totally bounded and closed implies compact??

Is there a fault in exercice 9.3.1 b) from Analysis by Zorich? The exercice asks to prove that a subset of a metric space is compact if and only if it is totally bounded and closed. But I have a counterexample for it: Consider the open unit ball $$B(0,1)$$ in $$\mathbb{R}^n$$ as a metric space itself. Then it is closed in itself and totally bounded, but not compact.

Am I right?

• " closed in itself " ? Commented Dec 18, 2018 at 14:48
• Are you suggesting $B(0,1)$ endowed with subspace topology of $\mathbb{R}^n$ ? Commented Dec 18, 2018 at 14:55
– Jiu
Commented Dec 18, 2018 at 14:56
• it is not closed in $\mathbb R^n$ which is presumably what is implied... Commented Dec 18, 2018 at 19:15
• @Carmeister but one does not write simply “metric space” and imply that it is $\mathbb{R}^n$.
– Jiu
Commented Dec 19, 2018 at 3:01

You are right.

For a subset $$A$$ of metric space $$(X,d)$$ to be compact, it is not enough that $$A$$ is totally bounded and closed (since $$X$$ is always closed). However, the correct assumptions to conclude that $$A$$ is compact are that it is totally bounded and complete. You can follow the link I included in my previous answer to a similar question to see that proof of this fact.

• Almost certainly, the problem should have been to prove that a subset of a complete metric space is compact iff it is closed and totally bounded. This is true because the subset will then be complete itself if and only if it is closed in $X$. I do not know if Zorich accidently left off the word "complete" in the exercise, or if Jiu just missed it in the context. Commented Dec 18, 2018 at 17:10
• @PaulSinclair That's also my understanding as well. Regardless, I decided to answer the question as is presented here since I don't have access to the book soI couldn't check. Commented Dec 18, 2018 at 17:18
• @PaulSinclair I don’t know how to upload a picture but there was no “complete” in the exercise. I typed the exercise word for word apart from “Show that”.
– Jiu
Commented Dec 18, 2018 at 23:22
• @Jiu - I'll take your word for it. You have to understand that people missing important words in their exercises is a fairly common occurrence here. Sometimes they fail to recognize that the word is important, and sometimes they just read past it. It happens to me, too.. So without access to the book, I did not have a way to verify that the exercise was indeed in error. But if you've double-checked, then I am sure that it is. Commented Dec 18, 2018 at 23:58

$$B(0,1)$$ is closed in the space $$B(0,1)$$ endowed with the subspace topology of $$\mathbb{R}^n$$. But this is true for every $$S \subset \mathbb{R}^n$$ if we endow $$S$$ with the subspace topology.

The requirement is that the set is closed in $$\mathbb{R}^n$$, not in itself.

Moreover, in a generic tolopogical space X, given $$A \subset X$$, the equivalence " $$A$$ is compact if and only if closed and totally bounded" is correct in the case the ambient space $$X$$ is complete. In this case every closed subspace of $$X$$ is also complete.

• The requirement in the exercise is that the set is closed in a metric space. Since every subspace of a metric space is a metric space itself, this is an empty requirement.
– Jiu
Commented Dec 19, 2018 at 3:04