Continuous Functions on Metric Spaces Review

# Continuous Functions on Metric Spaces Review

We will now review some of the recent material regarding continuous functions on metric spaces.

Let $(S, d_S)$ and $(T, d_T)$ be metric spaces and let $f : S \to T$.

• On the Continuity of Functions on Metric Spaces page we said that $f$ is continuous at $p \in S$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $d_S(x, p) < \delta$ then $d_T(f(x), f(p)) < \epsilon$.
• Equivalently, $f$ is continuous at $p$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that the image of the ball centered at $p$ with radius $\delta$ (in $S$) is fully contained in the ball centered at $f(p)$ with radius $\epsilon$ (in $T$), i.e.:
(1)
\begin{align} \quad f(B_S(p, \delta)) \subseteq B_T(f(p), \epsilon) \end{align}
• We noted that if $p$ is an accumulation point of $S$ then $f$ is continuous at $p$ if and only if:
(2)
\begin{align} \quad \lim_{x \to p} f(x) = f(p) \end{align}
• On the other hand, if $p$ is an isolated point of $S$ then $f$ is always continuous at $p$ since there will exist a $\delta > 0$ such that $B_S(p, \delta)$ contains only $p$, and so $f(B_S(p, \delta)) = \{ f(p) \}$ and $\{ f(p) \} \subseteq B_T(f(p), \epsilon)$ for all $\epsilon > 0$.
• On the Sequential Criterion for the Continuity of a Function on Metric Spaces page we looked at very important criterion for the continuity of a function on metric spaces at a point. We said that $f$ is continuous at $p$ if and only if for all sequences $(x_n)_{n=1}^{\infty}$ in $S$ that converge to $p$ we have that the sequences $(f(x_n))_{n=1}^{\infty}$ in $T$ converge to $f(p)$.
• On The Continuity of Composite Functions on Metric Spaces page we further showed that if $(S, d_S)$, $(T, d_T)$, and $(U, d_U)$ are all metric spaces with $f : S \to T$ and $g : T \to U$ then if $f$ is continuous at $p \in S$ and $g$ is continuous at $f(p) \in T$ then the composite function $g \circ f : S \to U$ is continuous at $p$.
• We then turned out attention towards the continuity of complex-valued functions. On the Continuity of Combinations of Complex-Valued Functions page we said that if $(S, d_S)$ and $(\mathbb{C}, d)$ are metric spaces where $d(x, y) = \mid x - y \mid$ for all $x, y \in \mathbb{C}$, and if $f, g : S \to \mathbb{C}$, $p \in S$, and $c$ is a constant then if $f$ and $g$ are continuous at $p$ then $f + g$, $f - g$, $cf$ (for any $c, and [[$ fg$are continuous at$p$. • We then looked at the continuity of vector-valued functions. On the Continuity of Combinations of Vector-Valued Functions page we said that if$(S, d_S)$and$(\mathbb{R}^n, d)$are metric spaces where$d(\mathbf{x}, \mathbf{y}) = \| \mathbf{x} - \mathbf{y} \|$for all$\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$, and if$\mathbf{f}, \mathbf{g} : S \to \mathbb{R}^n$,$p \in S$, and$\lambda$is a constant then if$\mathbf{f}$and$\mathbf{g}$are continuous at$p$then$\mathbf{f} + \mathbf{g}$,$\mathbf{f} - \mathbf{g}$,$\lambda \mathbf{f}$,$\mathbf{f} \cdot \mathbf{g}$, and$\| \mathbf{f} \|$are all continuous at$p$. • Similarly,$f$is continuous on$S$if and only if for every closed set$V$in$T$we have that the inverse image$f^{-1}(V)$is closed in$S$. • We then looked at a very nice property of continuous functions with compact domains. On the Continuous Functions on Compact Sets of Metric Spaces page. We saw that if$f$is continuous,$X \subseteq S$, and$X$is compact in$S$then the image$f(X)$is compact in$T$, and as a corollary,$f(X)$will be closed, bounded, and every infinite subset of$f(X)$will have an accumulation point. • On the The Extreme Value Theorem for Continuous Functions on Compact Sets of Metric Spaces page we looked at a generalization of the famous Extreme Value theorem. We said that$(S, d_S)$and$(\mathbb{R}, d)$are metric spaces where$d(x, y) = \mid x - y \mid$for all$x, y \in \mathbb{R}$, and if$X \subseteq S$is compact in$S$and$f : S \to T$is continuous then$f(X)$is closed and bounded in$T$and achieves its supremum and infimum on$X$, that is, there exists$p, q \in X\$ such that:
(3)
\begin{align} \quad f(p) = \sup \{ f(x) : x \in X \} \quad, \quad f(q) = \inf \{ f(x) : x \in X \} \end{align}