The proof is from Stephen Abbott's *Understanding Analysis, Second Edition*.

Theorem: A function that is continuous on a compact set $K$ is uniformly continuous on K.

Proof: Assume $f:K\to\Bbb{R}$ is continuous at every point of a compact set $K\subseteq\Bbb{R}$. To prove that $f$ is uniformly continuous on $K$ we argue by contradiction.

By the *Sequential Criterion for Absence of Uniform Continuity* (A function $f:A\to\Bbb{R}$ fails to be uniformly continuous on $A$ if and only if there exists a particular $\epsilon_0\gt0$ and two sequences $(x_n)$ and $(y_n)$ in $A$ satisfying $\vert x_n-y_n\vert\to0$ but $\vert f(x_n)-f(y_n)\vert\geq\epsilon_0$), if $f$ is not uniformly continuous, then there exist two sequence $(x_n)$ and $(y_n)$ in $K$ such that $\lim\vert x_n-y_n\vert=0$ while $\vert f(x_n)-f(y_n)\vert\ge\epsilon_0$ for some particular $\epsilon_0\gt0$. Because $K$ is compact, then sequence $(x_n)$ has a convergent subsequence $(x_{n_k})$ with $x=\lim x_{n_k}$ also in $K$.

We could use the compactness of $K$ again to produce a convergent subsequence of $(y_n)$, but notice what happens when we consider the particular subsequence $(y_{n_k})$ consisting of those terms in $(y_n)$ that correspond to the terms in the convergent subsequence $(x_{n_k})$. By the *Algebraic Limit Theorem*, $\lim(y_{n_k})=\lim((y_{n_k}-x_{n_k})+x_{n_k})=0+x$. The conclusion is that both $x_{n_k}$ and $(y_{n_k})$ converge to $x\in K$. Because f is assumed to be continuous at $x$, we have $\lim f(x_{n_k})=f(x)$ and $lim f(y_{n_k})=f(x)$. A contradiction arises when we recall that $x_n$ and $y_n$ were chosen to satify $\vert f(x_n)-f(y_n)\vert\geq\epsilon_0$ for all $n\in\Bbb{N}$.

My question: what contradiction?

Indeed, if by "consider the particular subsequence $(y_{n_k})$ consisting of those terms in $(y_n)$ that correspond to the terms in the convergent subsequence $(x_{n_k})$", the author were referring to a subsequence within the subsequence $y_{n_k}$ such that each $n_k$ in $y_{n_k}$ is equal to the $n_k$ in $(x_{n_k})$, then the proof clearly follows; but there is no guarantee that such a subsequence within $(y_{n_k})$ exists.

Please help. Many thanks in advance!

## Best Answer

Here is the contradiction: $$ \epsilon_0\le |f(x_{n_k}) - f(y_{n_k})|\le |f(x_{n_k}) - f(x)| + |f(x)-f(y_{n_k})|\,\to\,0 $$ as $k\to\infty$.

They never did, actually. They choose a convergent subsequence $x_{n_k}\to x$ and then conclude that also $y_{n_k}\to x$.