# Show that if $u_n:=p(f(x_n),g(y_n))$ for all $n\in \mathbb{N}$, then the sequence $(u_n)$ converges.

Let $$(X,d)$$ and $$(Y,p)$$ be metric spaces, where $$X$$ is complete, and define continuous functions $$f,g:X \to Y$$. If $$(x_n)$$ and $$(y_n)$$ are Cauchy sequences in $$X$$ and $$u_n:=p(f(x_n),g(y_n))$$ for all $$n\in \mathbb{N}$$, show that the sequence $$(u_n)$$ converges.

What I knew was:

1. Since $$f$$ is continuous, say at $$x_0\in X$$, then for any $$\varepsilon >0$$, there exists $$\delta _1>0$$ such that for all $$x\in X$$ with $$d(x,x_0)<\delta _1$$, we have $$p(f(x),f(x_0))<\varepsilon$$. Similarly, since $$g$$ is continuous, say at $$y_0\in X$$, then for any $$\varepsilon >0$$, there exists $$\delta _2>0$$ such that for all $$y\in X$$ with $$d(y,y_0)<\delta _2$$, we have $$p(f(y),f(y_0))<\varepsilon$$.
2. Since $$X$$ is complete, then $$(x_n)$$ and $$(y_n)$$ are convergent, say $$x_n\to x$$ and $$y_n\to y$$, where $$x,y\in X$$. This means that for any $$\varepsilon >0$$, there exists $$N_1,N_2\in \mathbb{N}$$ such that for all $$n\geq N_1$$ and $$n\geq N_2$$, we have $$d(x_n,x)<\varepsilon$$ and $$d(y_n,y)<\varepsilon$$, respectively.

I didn't know yet how to apply these facts to showing that $$(u_n)$$ is converges. Any helps? Thanks in advanced.

• Hint : $\tag*{}$ 1. Show that $X \times X$ is complete for any product metric.$\tag*{}$ 2. Show that the sequence $(x_n, y_n)$ is Cauchy and deduce that it converges. $\tag*{}$ 3. Show that $(z_1,z_2) \mapsto p(z_1,z_2)$ is continuous on $Y \times Y$. $\tag*{}$ 4. Deduce that $(t_1, t_2) \mapsto p(f(t_1),g(t_2))$ is continuous on $X \times X$. $\tag*{}$ 5. Using 2. and 4., deduce that $(u_n)$ converges. Jun 8, 2022 at 10:31
• @TheSilverDoe What about the answer below ? Does it correct? Jun 8, 2022 at 11:40

By completeness of $$X$$, $$(x_n)$$ and $$(y_n)$$ converge to some $$x$$ and $$y$$ in $$X$$ respectively. Now, for any $$\epsilon>0$$, there exists $$\delta>0$$ such that for any $$z$$ within $$\delta$$ of $$x$$, $$p(f(z),f(x))<\frac{\epsilon}{2}$$, and for any $$z$$ within $$\delta$$ of $$y$$, $$p(g(z),g(y))<\frac{\epsilon}{2}$$ by continuity of $$f$$ and $$g$$. Then, let $$N\in\mathbb{N}$$ such that for all $$n\geq N$$, $$d(x_n,x)<\delta$$ and $$d(y_n,y)<\delta$$. Then, the triangle inequality yields $$p(f(x_n),g(y_n))-p(f(x),g(y))\leq \big(p(f(x_n),f(x))+p(f(x),g(y))+p(g(y),g(y_n))\big)-p(f(x),g(y)),$$ and similarly, $$p(f(x),g(y))-p(f(x_n),g(y_n))\leq \big(p(f(x),f(x_n))+p(f(x_n),g(y_n))+p(g(y_n),g(y))\big)-p(f(x_n),g(y_n)),$$ so $$|p(f(x_n),g(y_n))-p(f(x),g(y))| \leq p(f(x_n),f(x))+p(g(y_n),g(y))<\epsilon$$ and thus $$p(f(x_n),g(y_n))\rightarrow p(f(x),g(y))$$.

• Wait,, does $f$ and $g$ continuous at the same point $x$? Jun 8, 2022 at 10:32
• As $f$ and $g$ are continuous, they are continuous at each point of $X$. I have used the facts that $f$ is continuous at the limit $x$ and $g$ is continuous at the limit $y$. If the maps are not continuous at these points, then $p(f(x_n),g(y_n))$ would not necessarily converge. Try forming a counterexample in this case by constructing non-constant sequences $x_n\rightarrow x$, $y_n\rightarrow y$, with some $f$ and $g$ that vary wildly around these points.
– Alex
Jun 8, 2022 at 10:44
• Using the notation in the question, perhaps it is more clear if we explicitly find separate $\delta_1$ for $f$ and $\delta_2$ for $g$. This is not a problem as we can simply take $\delta=\min(\delta_1,\delta_2)$. Similarly, we can fine separate $N_1$ and $N_2$ in $\mathbb{N}$ for the sequences $(x_n)$ and $(y_n)$, and again take $N=N_1+N_2$.
– Alex
Jun 8, 2022 at 11:05
Since $$X$$ is complete and both $$(x_n)$$ and $$(y_n)$$ are Cauchy sequences in $$X$$, then they are convergent in $$X$$. Let's say $$x_n \to x \in X$$ and $$y_n \to y \in X$$. Since both $$f$$ and $$g$$ are continuous in $$X$$, then they are continuous at every points in $$X$$. In particular, $$f$$ continuous at $$x$$ in $$X$$ and $$g$$ continuous at $$y$$ in $$X$$. Now, since $$x_n \to x$$ and $$f$$ continuous at $$x$$ in $$X$$, then $$f(x_n) \to f(x)$$. Similarly, $$g(y_n) \to g(y)$$. Claim: $$p(f(x_n),g(y_n))$$ converges to $$p(f(x),g(y))$$. Before we prove it, notice by definition of metric that $$p(f(x_n),g(y_n)) \le p(f(x_n),f(x)) + p(f(x),g(y)) + p(g(y),g(y_n))$$ or $$p(f(x_n),g(y_n)) - p(f(x),g(y)) \le p(f(x_n),f(x)) + p(g(y),g(y_n)), \qquad (1)$$ and $$p(f(x),g(y)) \le p(f(x),f(x_n)) + p(f(x_n),g(y_n)) + p(g(y_n),g(y)$$ or $$p(f(x),g(y)) - p(f(x_n),g(y_n)) \le p(f(x),f(x_n)) + p(g(y_n),g(y). \qquad (2)$$
Let $$\epsilon>0$$ be given. Since $$f(x_n) \to f(x)$$, there exists $$N_1 \in \Bbb N$$ such that for all $$n \ge N_1$$, we have $$p(f(x_n),f(x)) < \frac{\epsilon}{2}$$. Similarly, since $$g(y_n) \to g(y)$$, there exists $$N_2 \in \Bbb N$$ such that for all $$n \ge N_2$$, we have $$p(g(y_n),g(y)) < \frac{\epsilon}{2}$$. Let $$N=\max\{N_1,N_2\}$$. Then $$N \in \Bbb N$$ and for all $$n \ge N$$, we have \begin{align*} |p(f(x_n),g(y_n)) - p(f(x),g(y))| &\le p(f(x_n),f(x)) + p(g(y_n),g(y)) \qquad \text{(By (1) and (2))} \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ &= \epsilon. \end{align*} Hence, $$p(f(x_n),g(y_n))$$ converges to $$p(f(x),g(y))$$, as desired. Q.E.D.