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How could we show that the metric space

$$X=\{0\} \cup \{\frac{1}{n}:n \in \mathbb{Z} - \{0\}\}$$ with the metric it inherits as a subset of $\mathbb{R}$ is complete?

Thoughts

Complete metric spaces are those in which all Cauchy sequences converge to a point within the space. For any Cauchy sequence $(x_n)$ in the space, $|x_n|<1$ and so the sequence is bounded; bounded Cauchy sequences in $\mathbb{R}$ converge in $\mathbb{R}$ and so the limit, $x$ say, lies in $\mathbb{R}$.

Suppose that $x \notin X$. I imagine this leads to a contradiction but I can't see what it is.

Any help would be appreciated. Regards, MM.

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A subspace of a complete space is complete if and only if it is closed. See also completeness and closedness for a subset in a metric space. –  Martin Sleziak Jan 29 '12 at 16:18

5 Answers 5

up vote 4 down vote accepted

HINT (a variation of your approach):

Show that $X$ is a closed subset of a complete metric space. [In fact, the same argument shows that every closed and bounded subset of the reals is complete.]

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So, $\mathbb{R} \setminus X$ is open, and $\mathbb{R}$ is complete, so $(\mathbb{R} \setminus X)^{c}=X$ is a closed subset of $\mathbb{R}$ and thus complete. Is that right? –  Mathmo Jan 29 '12 at 16:50
    
That is correct. (I am not sure why :-), but I originally added boundedness condition on $X$. That is not needed.) –  Srivatsan Jan 29 '12 at 16:52
    
Is $\mathbb{R} \setminus X$ open because we can always find an interval between two numbers $\frac{1}{m}$ and $\frac{1}{m+1}$ in $\mathbb{R}$? i.e. $\mathbb{R} \setminus \{a_1,a_2,....., \}$ will be open for any countable number of singletons $a_n$ by the density of $\mathbb{R}$? –  Mathmo Jan 29 '12 at 16:57
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@MiamiMaths No, it is not true that any countable set $ \{a_1,a_2, \ldots \}$ is always closed in the reals. Two useful counterexamples to remember are (a) set of rationals, (b) $\{ \frac{1}{n} : n > 0 \}$. // But it is true that in your specific example, $\mathbb{R} \smallsetminus X$ can be written as the union of intervals of the form $\left( \frac{1}{m}, \frac{1}{m+1} \right)$ and $\left( -\frac{1}{m+1}, -\frac{1}{m} \right)$; hence $\mathbb{R} \smallsetminus X$ is open. –  Srivatsan Jan 29 '12 at 17:02
    
Thanks! Very useful. –  Mathmo Jan 29 '12 at 17:37

Hint: If $\{x_n\}$ is any sequence in $X$, it has a monotone subsequence. If this subsequence is not eventually constant, what must it converge to?

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@Srivatsan Thanks, somehow, I thought $\Bbb Z$ was $\Bbb N$. :) –  David Mitra Jan 29 '12 at 16:26
    
Nice fix. :-) I deleted my comment since it is now irrelevant and misleading. (+1, but I ran out of votes for today. I will vote it up tomorrow.) –  Srivatsan Jan 29 '12 at 16:29

I am not sure how much this will help, but it's an interesting approach nonetheless:

  1. Every compact metric space is complete (this is because every sequence must have a converging subsequence, and thus every Cauchy sequence is convergent).

  2. Every closed and bounded subset of $\mathbb R$ is a compact metric space.

  3. A set is closed if and only if it contains all its limit points.

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@MiamiMaths: Statement (2.) is the Heine-Borel theorem. –  Srivatsan Jan 29 '12 at 16:43
    
@Srivatsan: Yes. It is. –  Asaf Karagila Jan 29 '12 at 16:43

One way to tackle the problem is to show that the given set is closed. You will have the required result as closedness implies completeness in complete metric spaces and $\mathbb{R}$ is complete.

Observe that $X' = \{0\} $ Here $X'$ denotes the derived set of $X$ since the only limit point of $X$ is $0$. Given any other $1/n \in X $ choosing $ \delta = min \{|1/(n+1)-1/n|,|1/(n-1)-1/n| \} $ you can see that $(B(1/n, \delta)\setminus\{1/n\})\cap X = \emptyset. $ Thus $1/n \notin X'$ for all $n.$ Since $X' \subset X$ we have that $X$ is closed and hence it is complete.

If you want to tackle the question using the definition of a complete metric space, you must first observe that every Cauchy sequence in $X$ either becomes eventually constant or converges to $0$.

Obviously, it is easier to approach the problem using the previous method.

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The easiest approach is probably to show that $X$ is closed and use the fact that a closed subset of a complete metric space (with the subspace metric) is complete.

To see that $X$ is closed, there are many different ways. Here's one useful result:

If $U=\bigcup_{i\in I}U_{i}$ is a union of closed sets $U_{i}\subset Y$, so that for any $x\in Y\setminus U$ there exists a neighbourhood $B_{x}\subset Y$ such that $B_{x}\cap U_{i}\neq \emptyset$ for only finite amount of indices $i\in I$, then U is a closed subset of $Y$ (note that the index set $I$ can be arbitrary!).

In your case, write the space $X$ as a union of singletons (closed sets), and note that origin is the only point of reals which doesn't have the above property and that $0\in X$.

Or, simply show that $\mathbb{R}\setminus X$ is open by considering the two cases $|x|\leq 1$ and $|x|>1$ when $x\in \mathbb{R}\setminus X$.

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