Exercise 4.10

Question

\def\eps{\varepsilon}
Exercise 10:
    Complete the details of the following alternative proof of Theorem 4.19: If $f$ is not uniformly continuous, then for some $\varepsilon>0$ there are sequences $\{p_n\}$,
    $\{q_n\}$ in $X$ such that $d_X(p_n,q_n)\rightarrow 0$ but $d_Y\bigl(f(p_n),f(q_n)\bigr)>\varepsilon$. Use Theorem 2.37 to obtain a contradiction.

analambanomenos · Accepted Answer

\def\eps{\varepsilon}

By Theorem 2.37, $\{p_n\}$ has a subsequence which converges to $p\in X$. Replace $\{p_n\}$ with this subsequence and replace $\{q_n\}$ with the corresponding
    subsequence. Similarly, $\{q_n\}$ has a subsequence which converges to $q\in X$. Again replace $\{q_n\}$ with this subsequence and replace $\{p_n\}$ with the
    corresponding subsequence. Since $d_X(p_n,q_n)$ converges to 0, we must have $p=q$. Hence by continuity, $f(p_n)$ and $f(q_n)$ must both converge to $f(p)=f(q)$,
    contradicting the assumption that $d_Y\bigl(f(p_n),f(q_n)\bigr)>\varepsilon$.

Exercise 4.10

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Exercise 4.10

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