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The question I've been given is as follows:

Given that $(X,d),(Y,e)$ are complete metric spaces, show that the metric space $(M,f)$ is complete where $M=X \times Y$ and $f:M \times M\rightarrow \mathbb{R}$ is defined by $f((x_1,y_1),(x_2,y_2))=d(x_1,x_2)+e(y_1,y_2)$

The solution I have produced is:

Since $(X,d)$ is complete we must have that for any Cauchy sequence $(x_n)$ in $X,$ $\exists x \in X$ where $d(x_n,x)\rightarrow 0$ as $n\rightarrow \infty$,
Similarly, for any Cauchy sequence $(y_n)$ in $Y$, $\exists y \in Y$ where $e(y_n,y)\rightarrow 0$ as $n \rightarrow \infty$

Let $A_n=(x_n,y_n)$ be a sequence in $M$, then $$\lim_{n\to\infty}f((x_1,y_1),(x,y))=\lim_{n\to\infty}d(x_n,x)+\lim_{n\to\infty}e(y_n,y)=0$$Also, note that since $x\in X$, $y\in Y$, we have that $(x,y)\in M=X\times Y$.
So the (arbitrary) sequence $A_n$ converges to its limit which lies in M, therefore the metric space $(M,f)$ is complete.


Looking over this proof now I see that I haven't stated anywhere that the sequence $A_n$ is Cauchy, would this answer nonetheless be likely to achieve full credit? More specifically, is the fact that $A_n$ is Cauchy induced from $(x_n)$ and $(y_n)$ being Cauchy or would I need to prove this rigorously?

  • Your doubts are right, the solution is wrong. Not every sequence in $X\times Y$ converges, just as not any sequence in $X$ or $Y$ is Cauchy. You have to prove that any Cauchy sequence in $X\times Y$ converges to some limit. So your solution has to start from "let $A_n$ be an arbitrary Cauchy sequence in $X\times Y$", and you have to prove it has some limit. You on the other hand started from considering some $x_n$ and $y_n$, and building $A_n$ from them. No, you need to take a general Cauchy sequence. Then from it you get sequences $x_n$ and $y_n$, and use what you know about $X, Y$. – Mark Apr 05 '23 at 16:07
  • Okay. If I am assuming $A_n$ as I defined it is Cauchy, would I then just need to prove that, with that assumption, we get that $x_n$ and $y_n$ are also Cauchy and that, due to completeness, their limits of convergence lie in $X$ and $Y$ respectively? –  Apr 05 '23 at 16:14
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    Yes. Define $A_n$ to be a Cauchy sequence, and write $A_n=(x_n, y_n)$. Show that $x_n$ and $y_n$ are Cauchy (this is easy, but still requires a proof), and using the completeness of $X$ and $Y$ they converge to some $x\in X$ and $y\in Y$ respectively. Finally, show that $A_n\to (x,y)$. – Mark Apr 05 '23 at 16:17

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As pointed out in the comments, you solution does not work, as you start with considering Cauchy sequences in $X$ and $Y$ separately, and from that construct a sequence in $X\times Y$, however what you need to do it start with a Cauchy sequence in $X\times Y$. So what you should do is start with a Cauchy sequence in $X\times Y$, show that the component sequences are Cauchy, and then invoke the completeness of $X$ and $Y$ to get the result. I include a proof below which you can have a look at once you've tried this.

Let $\{(x_j,y_j)\}_{j\in\mathbb{N}}$ be a Cauchy sequence in $X\times Y$. This means that for any $\varepsilon>0$ we can find an $N\in\mathbb{N}$ such that

$$f((x_i,y_i),(x_j,y_j))<\varepsilon$$

for all $i,j\geq N$. But notice also then that if $i,j\geq N$, then

$$d(x_i,x_j)=f((x_i,y_i),(x_j,y_j))-e(y_i,y_j)\leq f((x_i,y_i),(x_j,y_j))<\varepsilon$$

and similarly

$$e(y_i,y_j)<\varepsilon.$$

This means that $\{x_j\}_{j\in\mathbb{N}}$ and $\{y_j\}_{j\in\mathbb{N}}$ are Cauchy, and so they converge to some $x\in X$ and $y\in Y$ respectively due to completeness. It follows that

$$\lim_{j\to\infty}f((x_j,y_j),(x,y))=\lim_{j\to\infty}\biggl(d(x_j,x)+e(y_j,y)\biggr)=0,$$

and so $(x_j,y_j)\to(x,y)$ as $j\to\infty$. As $\{(x_j,y_j)\}_{j\in\mathbb{N}}$ was arbitrary, it follows that $(X\times Y,f)$ is complete.

Lorago
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