# Sets where Heine-Borel theorem works [duplicate]

By Heine-Borel theorem, a closed and bounded subset of the Euclidean space is compact. If we analyze the proof, the only characteristic of Euclidean space that we need is: every bounded subset is contained in a compact subset. Is there a special name this kind of sets?

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## marked as duplicate by Alex Becker♦, Nate Eldredge, Asaf Karagila, t.b., Zhen LinMar 31 '12 at 0:04

I remember my teacher telling us they were Heine Borel Spaces. Let me look up before I write an answer. –  user21436 Mar 11 '12 at 13:06
I think HB and this property are equivalent and hence I'll write an answer. –  user21436 Mar 11 '12 at 13:14
@Alex How can something that asks for a name and that asks you some examples be duplicates? –  user21436 Mar 11 '12 at 13:27
@KannappanSampath The question and answers answer the OP's question fully IMHO. –  Alex Becker Mar 11 '12 at 13:28
I think OP should clarify if (s)he wants to know the name of those spaces in which every bounded set is contained in a compact set or just a list of Heine Borel spaces! Please respond. –  user21436 Mar 11 '12 at 13:46

This proves that HB is equivalent to the fact that every bounded subset is a subset of a compact set. Clearly, I am doing it on a metric space, only then this makes sense. As pointed out OP seems to understand this equivalence and hence is irrelevant.

Heine-Borel Property:

A metric space $(X, d)$ is said to be Heine-Borel if any closed and bounded subset of it is compact.

You'd like to know the name of those spaces in which a bounded set is contained in a compact set (BIC).

We claim that this property is equivalent to the Heine-Borel (HB) Property.

• $(BIC\implies HB)$ Suppose a space $X$ is $BIC$. Let $B$ bounded subset of $X$. Because $X$ is $BIC$, $B \subseteq K$ where $K$ is compact. If $B$ is also closed, you have closed subset of a compact set which is compact. Hence the space is $HB$.
• $(HB \implies BIC)$ Let $B$ bounded subset of $X$ that has $HB$. We need to find a compact subset $K$ such that $B \subseteq K$. Then the closure, $\bar B$ is closed and bounded, hence compact, because of $HB$ property and note that $B \subseteq \bar B$. Hence, $X$ is $BIC$.

So, we can as well call these Heine-Borel spaces.

In fact, another characterisation of Heine Borel spaces is that, bounded sets are also totally bounded.

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You really need to specify that you’re talking about metric spaces, so that bounded is meaningful. Also, in the second half of the argument you seem to think that if $B$ is not closed, it must be open, which isn’t true. You might as well just go directly to $\operatorname{cl}B$ in the first place and not split the argument into cases. But I suspect that the whole answer is superfluous: according to the question, joh already understands this equivalence. –  Brian M. Scott Mar 11 '12 at 19:47
@Brian Doh. I am sorry then. This answer is irrelevant. :-( –  user21436 Mar 11 '12 at 20:22
@BrianM.Scott I have made it CW because it may be of some help to other users, sometime later. –  user21436 Mar 11 '12 at 20:28