Continuous Functions

The notion of continuity is based on that of limit so we define that for functions of arbitrary metric spaces first.

Definition. Let $(X, \rho)$ and $(Y, \sigma)$ be metric spaces and $f: E \to Y$ with $E \subseteq X$ be a function and $x_0$ be a limit point of $E$ and $y_0 \in Y$. Then $f(x) \to y_0$ as $x \to x_0$ or $\lim_{x \to x_0} f(x) = y_0$ if

\[(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in E : 0 < \rho(x, x_0) < \delta) \; \sigma(f(x), y_0) < \varepsilon\]

The point $x_0$ doesn’t have to be in the domain of the function.

Proposition. Let $(X, \rho)$ and $(Y, \sigma)$ be metric spaces and $f: E \to Y$ with $E \subseteq X$ be a function and $x_0$ be a limit point of $E$ and $y_0 \in Y$. Then $f(x) \to y_0$ as $x \to x_0$ iff for every sequence $(x_n)$ in $E - \Set{x_0}$ such that $x_n \to x_0$, $f(x_n) \to y_0$ as $n \to \infty$.

Proof.

($\Rightarrow$) Suppose that $f(x) \to y_0$ as $x \to x_0$. Given $\varepsilon > 0$, there exists $\delta > 0$ such that

\[0 < \rho(x, x_0) < \delta \implies \sigma(f(x), y_0) < \varepsilon\]

On the other hand, for a sequence such that $x_n to x_0$ as $n \to \infty$, there exists $N > 0$ such that for all $n > N$, $\rho(x_n, x_0) < \delta$ and therefore $f(x_n) \to y_0$ as $n \to \infty$.

($\Leftarrow$) Suppose that whenever $x_n \to x_0$, $f(x_n) \to y_0$. Assume $f(x) \not\to y_0$ as $x \to x_0$, then there exists $\varepsilon > 0$ such that for all $\delta > 0$, $\rho(x, x_0) < \delta$ but $\sigma(f(x), y_0) \ge \varepsilon$. Therefore, there exists sequence $(x_n)$ such that for sufficiently large $n$, $\rho(x_n, x_0) < 1/n$ but $\sigma(f(x_n), y_0) \ge \varepsilon$ so $f(x_n) \not\to y_0$ as $n \to \infty$ which is a contradiction.

For normed vector space, the algebra of elements allows the usual algebra of limits, i.e.

\[\lim_{x \to x_0} a(x) f(x) + b(x) g(x) = ay_0 + bz_0\]

Continuity

We can now naturally extend the definition of continuity to functions of metric spaces.

Definition. Let $(X, \rho)$ and $(Y, \sigma)$ be metric spaces and $f: X \to Y$ be a function. Then $f$ is continuous at the point $x_0 \in X$ if

\[(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in X: \rho(x, x_0) < \delta) \; \sigma(f(x), f(x_0)) < \varepsilon\]

(or $f(B_\delta(x_0)) \subset B_\varepsilon(f(x_0))$).

The definition implies that $f$ is continuous at a limit point $x_0$ if $f(x) \to f(x_0)$ as $x \to x_0$. Also, $f$ is continuous at all isolated points of $X$ since there is always $\delta$ such that $x_0$ is the only point in $B_\delta(x_0)$.

Similarily, continuity can be characterized by sequences just like limits using similar proof.

Proposition. $f: X \to Y$ is continuous at the point $x_0$ iff for every sequence $(x_n)$ in $X$ such that $x_n \to x_0$, $f(x_n) \to f(x_0)$ as $n \to \infty$.

From the above we can see that continuity is a topological property which we can also characterized that by open/closed sets.

Proposition. $f: X \to Y$ is continuous on $X$ iff

  • whenver $G$ is open in $Y$, then $f^{-1}(G)$ is open in $X$; or

  • whenver $F$ is closed in $Y$, then $f^{-1}(F)$ is closed in $X$; or

Proof.

($\Rightarrow$) Suppose that $f$ is continuous and $G$ is open in $Y$. If $f^{-1}(G)$ is empty then it is open. Let $x_0 \in f^{-1}(G)$ so that $f(x_0) \in G$, then there exists $\varepsilon$ such that $B_\varepsilon(f(x_0)) \subset G$. Since $f$ is continuous, there exists $\delta$ such that $\sigma(f(x_0), f(x)) < \varepsilon$ whenever $\rho(x_0, x) < \delta$, therefore $f(x) \in G$ and $x \in f^{-1}(G)$ so $B_\delta(x_0) \subset X$ and $x_0$ is an interior point. Hence, $f^{-1}(G)$ is open.

($\Leftarrow$) Suppose that $G$ is open implies $f^{-1}(G)$ is open. Given $x_0 \in X$ and $\varepsilon > 0$. Since $B_\varepsilon(f(x_0))$ is open in $Y$, $f^{-1}(B_\varepsilon(f(x_0)))$ is open in $X$. Therefore, there exists $\delta$ such that $x \in B_\delta(x_0)$ then $f(x) \in B_\varepsilon(f(x_0))$ which implies $f$ is continuous.

The case for closed set can be derived from complements, i.e. if $E$ is any set, $f^{-1}(E) \cap f^{-1}(E’) = \emptyset$ and $f^{-1}(E) \cup f^{-1}(E’) = X$.

Note that it is not true that $S$ being open in $X$ implies $f(S)$ being open in $Y$.

In general for a bijective continuous function $f$, its inverse $f^{-1}$ needs not be continuous. We have a special notion if that is the case.

Definition. A continuous function $f: X \to Y$ is called a homeomorphism if its inverse $f^{-1}$ is also continuous. The metric spaces $X$ and $Y$ are said to be homeomorphic if there exists a homeomorphism between them.

By the above theorem we can see that if two metric spaces are homeomorphic then there is a bijection between the open (respectively closed) sets in the two spaces.

Definition. A bijection $f: X \to Y$ such that $\sigma(f(x_1), f(x_2)) = \rho(x_1, x_2)$ for all $x_1, x_2 \in X$ is called an isometry. The metric spaces $X$ and $Y$ are said to be isometric if there eixsts an isometry between them.

Uniform Continuity

Similar to pointwise convergence and uniform convergence of functions, we can do the same to define uniform continuity.

Definition. [Uniform Continuity] Suppose that $f: X \to Y$ is a function of metric spaces $X$ and $Y$. Then $f$ is said to be uniformly continuous on $X$ if

\[(\forall \varepsilon > 0)(\exists \delta)(\forall x_1, x_2 : \rho(x_1, x_2) < \delta) \; \sigma(f(x_1), f(x_2)) < \varepsilon\]

References