Exercise 1.5.12

Question

Let $\left(X, d_{	ext{disc}}ight)$ be a metric space with the discrete metric $d_{	ext{disc}}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $X$ is always complete.
\item When is $X$ compact, and when is $X$ not compact?
\end{enumerate}

Prakash Anand · Accepted Answer

(a) \begin{proof} Let $\left(x^{(n)}ight)_{n=m}^{\infty}$ be a Cauchy sequence in $X$ with respect to $d_{	ext {disc }}$. Then we know for every $\epsilon>0$ there exists an $N \geq m$ such that $d\left(x^{(j)}, x^{(k)}ight)<\epsilon$ for all $j, k \geq N$. However, since we are in the discrete metric, for this to be true, we must eventually have $N$ large enough so that $d\left(x^{(j)}, x^{(k)}ight)=0 \forall j, k \geq N$. So $x^{(j)}=x^{(k)}$ (that is, eventually we are tapping on the same spot over and over again), but each of the $x^{(j)}$ 's are in $X$, thus we are convergent in $X$. Therefore $X$ is complete (with respect to $d_{	ext {disc }}$ ).
\end{proof}

(b) $X$ is compact whenever $X$ is finite (this is clear since any sequence must contain a limit point (as there are only a finite number of places to go), and hence a convergent subsequence). $X$ is not compact whenever $X$ is infinite. Take any point in $X$, make it $x_{1}$, then a different point $x_{2}$, and we can do this forever without repeating points, and since each point is different the distance between each point will be 1, hence we have no limit points, and thus no convergent subsequences.

Exercise 1.5.12

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Exercise 1.5.12

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