I know that every uniformly continuous function is also Cauchy-continuous. But the converse is not always true.
Are there any specific conditions when Cauchy continuity implies uniform continuity ? Wikipedia says :
For $f : X \to Y $, if the domain $X$ is totally bounded, then every Cauchy-continuous function is uniformly continuous.
But I am having trouble understanding how this holds true. Can anyone please give a proof of this theorem ?
Also, it would be extremely helpful if someone could provide me with a general intuition for this result.
Best Answer
Here's the motivation for the proof:
When trying to prove uniform continuity, it's often useful to suppose otherwise and try to get a contradiction. What happens is you get two sequences $(x_n)$ and $(y_n)$ in $X$ which "get close to each other eventually", but $(f(x_n))$ and $(f(y_n))$ don't get close to each other in $Y$. You can often get a contradiction from this. In our case, we can use the total boundedness of $X$ to ensure that $(x_n)$ and $(y_n)$ are Cauchy, and then stitch them together to get a Cauchy sequence $x_1,y_1,x_2,y_2,\dots$ in $X$ such that $f(x_1),f(y_1),f(x_2),f(y_2),\dots$ is not Cauchy in $Y$, since $f(x_n)$ and $f(y_n)$ stay far apart when $n$ gets large.
Okay, here's the actual proof:
Let $X,Y$ be metric spaces and suppose that $f:X\to Y$ is Cauchy continuous but not uniformly continuous. Then there is some $\varepsilon>0$ such that, for all $\delta>0$ there are $x$ and $y$ in $X$ such that $d(x,y)<\delta$ but $d(f(x),f(y))>\varepsilon$. So, we can form two sequences $(x_n)$ and $(y_n)$ in $X$ such that, for all $n$, $d(x_n,y_n)<\frac{1}{n}$ and $d(f(x_n),f(y_n))>\varepsilon$. Since $X$ is totally bounded, $(x_n)$ has a Cauchy subsequence $(x_{m(n)})$. Since $d(x_{m(n)},y_{m(n)})\to 0$, the sequence $(z_n)$ given by $$z_1=x_{m(1)},z_2=y_{m(1)},z_3=x_{m(2)},z_4=y_{m(2)},\dots$$ is Cauchy. However, for all $n$, $d(f(z_n),f(z_{n+1}))>\varepsilon$, so $(f(z_n))$ is not Cauchy, which contradicts the Cauchy-continuity of $f$.