Proving part of the theorem that a continuous function on the bounded interval is uniformly continuous

Question

I am asked to prove the following. Suppose $D$ is a closed and bounded subset of $R$ and suppose $f:D\to R$ is continuous on $D$. Then $f$ is uniformly continuous: $ $Proof: Suppose by contradiction that $\{x_n\}$ and $\{y_n\}$ are sequences in $D$ s.t. $$ \lim_{n\to\infty} [x_n-y_n] =0$$

and the sequence $[f(x_n)-f(y_n)]$ does not converge to $0$.

We want to negate the statement of convergence for the above s.t. we have: "There exists an $\epsilon>0$ s.t. for all natural number k there exists an $n_k>k$ s.t. |$f(x_{n_k})-f(v_{n_k})$|$\geq \epsilon$.

Now apply the sequential compactness theorem to $\{x_{n_k}\}$ which is a subset of $D$. There is a limit point $x_0$ s.t. there is a further subsequence $\{x_{n_{k_j}}\}$ s.t. the limit as j tends to infinity is $x_0$, which is in $D$.

Now here's what I am supposed to prove: By supposing that $\{x_n\}$ and $\{y_n\}$ are sequences satisfying the fact that the image of a continuous function on a closed bounded interval is bounded above and assuming $\{x_n\}$ converges to $x$ I want to show that the limit as $j$ tends to infinity of $y_{n_{k_j}}$ is $x_0$. How would you do this part?

So basically I am just proving this one last part of the proof.

Answer 1

Suppose not, $\exists \{x_{n}\},\{y_{n}\} \in D$,s.t. $x_n-y_n \to 0$,but $|f(x_{n})-f(y_{n})|\geq \epsilon_0$. Since $D$ is bounded, we have convergent subsequence $\{x_{n_k}\}\to x_0,\{y_{n_k}\} \to y_0$ (Here you may think about why we can assume these two convergent subsequences have the same index).

Since $D$ is closed, $x_0, y_0\in D$. Since $x_{n_k}-y_{n_k}\to 0$, by uniqueness of limit, $x_0=y_0$ but $|f(x_{n_k})-f(y_{n_k})|\geq \epsilon_0$. which contradicts to continuity of $f$ at $x_0$.

Answer 2

Here is other proof maybe this would help you

If not, there is a $\epsilon>0$ and points $x_n,y_n\in D$, s.t. $d(x_n,y_n)<1/n$ and $d(fx_n,fy_n)\ge \epsilon$ for all $n$. Since any sequence in $D$ has a convergent subsequence, we may assume $x_n\to x$, for $x\in D$, so $y_n\to x$. By continuity at $x$, for $n$ large enough $fx_n,fy_n\in B(fx,\epsilon/2)$. Thus $d(fx_n,fy_n)<\epsilon$, contradiction.