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:
- 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 $p(f(y),f(y_0))<\varepsilon$.
- 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.
Best Answer
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 $p(f(x_n),g(y_n))\rightarrow p(f(x),g(y))$.