Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm stuck on how to show one of these is compact, and I want to verify my method for the other.

This one I am stuck on:

Proposition Let $(\mathbb{Q},d)$ be a metric space with $d(a,b)=|a-b|$. Let $E \subset \mathbb{R}$ and let $E=\{r \in \mathbb{Q}: r^2 < 2\}$. Show $E$ is not compact.

This one I believe I have solved and want to verify:

Proposition Let $\{x_n\}$ be a convergent sequence in $\mathbb{R}$ that converges to $x$. Let $E=\{x_n : n=1,2,3...\}\cup \{x\}$. Then $E$ is compact.

I believe I demonstrated this by the following. Since $\{x_n\}$ is convergent fix $\epsilon =1$ and then for $N \in \mathbb{N}$, there exists an $n \geq N$ such that $|x_n-x|<1$. Then take $\bar{\epsilon}=\max\{d(x_1,x)...d(x_N,x), 1\}$. Clearly $E \subset (\bigcup\limits_{k=1}^{N} N(x_k, \bar{\epsilon}))\bigcup N(x, 1))$ - this is a finite subcover of $N+1$ sets.

share|cite|improve this question
up vote 1 down vote accepted

For the first proposition, Heine-Borel Theorem asserts that every compact subset of $\mathbb{R}$ is closed and bounded. $E$ is clearly bounded, but is it closed? Can you find a limit point of $E$ that is not in $E$? Can you then find a Cauchy sequence in $E$ that does not have a limit in $E$?

The proof of the second proposition is incorrect. Why are you fixing $\epsilon$? The requirement for compactness is that whenever I give you a list of open sets whose union covers $E$, you can give me a finite subcover. I am not going to be so nice to only give you open sets expressible as open intervals of length $\bar{\epsilon}$....

Let me give you a hint: if $O$ is an open subset of $\mathbb{R}$ such that $x\in O$, then there are only finitely many of the $(x_n)$ that lies outside of $O$. (Firstly, why is that true? Secondly, how does this help you with the proof?)

share|cite|improve this answer
The first $N$ (finitely many) terms of a convergent sequence are more than $\epsilon$ apart. So my goal was to center a neighborhood around those and the limit point. So following your hint. Given $x \in O \subset \mathbb{R}$, this contains infinitely many points of the sequence. All that is left are the first $N$ terms. Could I just cover each one with an arbitrary open set (not defining a radius) centered at $x_n, n=1,2,...,N$? Also I noticed that my problem as stated is incorrect. I needed to show E is not compact. I have show that E is closed and bounded AND open. – emka Oct 25 '12 at 7:51
@abet: again, you need to pick a finite subcover from the cover that I gave you. So you cannot "just cover each one with an arbitrary open set", because said "arbitrary open set" may not belong to the cover that I gave you to start with. Now, the question is: how do you go about choosing a collection of open sets from the cover that I gave you, so that every one of the first $n$ points lives inside their union? – Willie Wong Oct 26 '12 at 7:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.