Exercise 18.1

Question

Prove that for functions $f\colon \mathbb{R} 	o \mathbb{R}$, the $\epsilon$-$\delta$ definition of continuity implies the open set definition.

Amit Awasthi · Accepted Answer

\begin{remark} Recall that $f$ is continuous if for every $\epsilon > 0$ and $x_0 \in \mathbb{R}$, there exists a $\delta > 0$ such that $|f(x) - f(x_0)| < \epsilon$ for all $x \in \mathbb{R}$ such that $|x-x_0| < \delta$. \end{remark} \begin{proof} Consider $x_0 \in \mathbb{R}$, and a corresponding neighborhood $V$ of $f(x_0)$; we then have $V \supseteq (f(x_0)-\epsilon,f(x_0)+\epsilon)$ for some $\epsilon > 0$ since $V$ is open. Then, by hypothesis there exists a $\delta > 0$ such that $f(x) \in (f(x_0)-\epsilon,f(x_0)+\epsilon)$ for all $x \in \mathbb{R}$ such that $x \in (x_0-\delta,x_0+\delta) = U$, which is open. Thus, $f(U) \subseteq V$, and so $f$ is continuous by Theorem $18.1$. \end{proof}

Dan Whitman · Answer

\def\e{\epsilon} \def\d{\delta} Recall that that $f: {\mathbb{R}} o {\mathbb{R}}$ is continuous at a point $x \in {\mathbb{R}}$ if, for every real $\e > 0$, there is a real $\d > 0$ such that $\left|f(y) - f(x) ight| < \e$ for every real $y$ where $\left|y - x ight| < \d$. We say that $f$ itself is continuous if it is continuous at every $p \in {\mathbb{R}}$. \begin{proof} Suppose that $f: {\mathbb{R}} o {\mathbb{R}}$ is continuous by the $\e$-$\d$ definition above. We show that this implies the open set definition by showing that $f$ satisfies (4) in Theorem~18.1. So consider any $x \in {\mathbb{R}}$ and any neighborhood $V$ of $f(x)$. Then of course there is a basis element $(c,d)$ containing $f(x)$ such that $(c,d) \subset V$. Let $\e = \min(f(x)-c, d-f(x))$, noting that $\e > 0$ since $c < f(x) < d$. It is then trivial to show that $(f(x)-\e, f(x)+\e) \subset (c,d) \subset V$ and contains $x$. Then, since $f$ is continuous at $x$, there is $\d > 0$ such that $\left|y - x ight| < \d$ implies that $\left|f(y) - f(x) ight| < \e$ for any real $y$. Let $U = (x-\d, x+\d)$, which is clearly a neighborhood of $x$. Now consider any $z \in f(U)$ so that $z = f(y)$ for some $y \in U$. Then we have that $x - \d < y < x+ \d$ so that clearly $-\d < y - x < \d$, from which it follows that $\left|y-x ight| < \d$. We then know that $\left|z - f(x) ight| = \left|f(y) - f(x) ight| < \e$ since $f$ is continuous. Hence $-\e < z - f(x) < \e$ so that $f(x) - \e < z < f(x) + \e$, and thus $z \in V$ since $(f(x)-\e, f(x)+\e) \subset V$. Since $z \in f(U)$ was arbitrary, this shows that $f(U) \subset V$, which shows that (4) holds for $f$ since $x$ was also arbitrary. \end{proof}

Exercise 18.1

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Exercise 18.1

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