$(X,d)$ is compact if each $(X,d_k)$ is compact

Question

For each $k\in \mathbb N$, let $d_k$ be a metric on $X$ such that $d_k(x,y)\leq 1$ for all $x,y\in X$. Define a new metric, $$d(x,y)=\sum_{k\geq 0}2^{-k}d_k(x,y)$$ Now show that if $(X,d_k)$ is compact for every $k$ then so is $(X,d)$.

Here is my attempt - let $(x_n)$ be an infinite sequence, I have extracted, through a diagonal argument, a subsequence $(x_{nn})$ which converges to, say, $y_k$ in the metric $d_k$, for every $k\in \mathbb{N}$. Now I do not understand how to proceed from here.

This question may have been already asked before, but I could not find it.

Answer 1

I will assume that $(X,d)$ is complete because otherwise the result is false. Suppose $(x_n)$ is a sequence in $(X,d).$ Then, there is a subsequence of $(x_n)$, conveniently written $(x_{1m})$, such that $x_{1m}\to x_1\in (X,d_1).$ There is a now a subsequence $(x_{2m})$, of $(x_{1m})$ such that $(x_{2m})\to x_2\in (X,d_2)$ and of course $(x_{2m})\to x_1\in (X,d_1)$ because it is a subsequence of $(x_{1m}).$

Continuing in this way, we obtain an array $(x_{nm})$ with the following properties:

$1).\ (x_{nm})\ \text {is a subsequence of}\ (x_{n'm}) \text{ whenever}\ n>n'$

$2).\ (x_{jm})\to x_j\in (X_j,d_j)\ \text{for all } 1\le j\le n$

Consider the diagonal of the array, i.e $(x_{nn})$ and let $m\in \mathbb N.$ Then, if $n\ge m$, the sequence $(x_{nn})$ is a subsequence of the sequence $(x_{mm},x_{m(m+1)},x_{m(m+2)},\cdots),\ $ which is a tail of the sequence $(x_{mk})$ and so converges to $x_m\in (X_m,d_m).$

Now consider $d(x_{nn},x_{mm})=\sum_{k\geq 0}2^{-k}d_k(x_{nn},x_{mm}).$ Since $d_k<1$ for all $k$, we can choose $N$ large enough so that for $\epsilon>0,$

$3). d(x_{nn},x_{mm})\le \sum^N_{k=0}2^{-k}d_k(x_{nn},x_{mm})+\frac{\epsilon }{2}.$

For each $0\le k\le N$ there is an $N_k$ such that if $n,m>N_k,$ then $d_k(x_{nn},x_{mm})<\frac{\epsilon }{2N}$ so it suffices to take $N>\max {N_k}$ to see that if $n,m>N$ then

$4). \sum^N_{k=0}2^{-k}d_k(x_{nn},x_{mm})<N\cdot \frac{\epsilon }{2N}=\frac{\epsilon }{2}.$

Combining $3).$ and $4).$, we have now that $(x_{nn})$ is Cauchy and so $x_{nn}\to x\in (X,d).$

Answer 2

The result is not true unless you have some compatibility conditions on $d_k$. The problem being $(X,d)$ need not be complete.

Example: Let $X=[0,1]$, and let $\rho$ be the usual metric, $\rho'$ is the metric obtained from $d$ swapping $0,1$, i.e., $\rho'(x,y)=\rho(f(x),f(y))$ where $$ f(t):= \begin{cases} 1 & t=0\\ 0 & t=1\\ t & \text{otherwise} \end{cases} $$ Clearly $(X,\rho)$ and $(X,\rho')$ are compact. Let $d_{odd}=\rho$ and $d_{even}=\rho'$. Then the sequence $x_n=1/n$ is Cauchy but it doesn't converge in $d$. $\Rightarrow\Leftarrow$

However, if $X$ is $d$-complete (e.g., if $d_k$ are all topologically equivalent), then the result holds, since

Claim: Every sequence has a $d$-Cauchy subsequence.

Gist of Proof: Impose a constraint on how quickly subsequence converges at each $d_k$ and pass to diagonal subsequence. For example, $d_k(x_{n_k(m)},x_{n_k(m')})<2^{-\min(m,m')}$, which implies a similar bound for $d$ when you pass to the diagonal subsequence.