Definitions and examples
Definition (Limit of a function between metric spaces)
Let \((X, d), (Y, \rho)\) be metric spaces, \(f : X \to Y\), \(a \in X\) and \(\ell \in Y\). We say that \(\lim_{z \to a} f(z) = \ell\) if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that
\[0 < d(z, a) < \delta \implies \rho(f(z), \ell) < \varepsilon.\]Note that to define the limit it is not necessary that \(f\) be defined at \(a\).
Definition (Continuous function)
The function \(f : X \to Y\) is continuous at \(a \in X\) if \(\lim_{z \to a} f(z) = f(a)\). That is, for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that
\[d(z, a) < \delta \implies \rho(f(z), f(a)) < \varepsilon,\]or equivalently, \(f(B_{X}(a, \delta)) \subseteq B_{Y}(f(a), \varepsilon)\). We simply say that \(f\) is continuous if it is so at every point of \(X\).
Definition (Lipschitz function)
\(f : X \to Y\) is \(\lambda\)-Lipschitz if for all \(x, y \in X\),
\[\rho(f(x), f(y)) \leq \lambda\, d(x, y).\]We simply say that \(f\) is Lipschitz if there exists \(\lambda \geq 0\) with this property.
Example (The distance to a point is \(1\)-Lipschitz)
Let \((X, d)\) be a metric space and \(a \in X\). The function \(f(x) = d(x, a)\) is \(1\)-Lipschitz, since from the triangle inequality
\[|d(x, a) - d(y, a)| \leq d(x, y),\]| that is $$ | f(x) - f(y) | \leq d(x, y)\(. In particular\)f$$ is continuous. |
Topological characterisations of continuity
Lemma (Continuity and preimages of open sets)
If \(f : X \to Y\) is continuous and \(G \subseteq Y\) is open, then \(f^{-1}(G)\) is open in \(X\).
Proof: Let \(x_{0} \in f^{-1}(G)\) and \(y_{0} = f(x_{0}) \in G\). Since \(G\) is open, there exists \(\varepsilon > 0\) with \(B_{Y}(y_{0}, \varepsilon) \subseteq G\). By continuity of \(f\) at \(x_{0}\), there exists \(\delta > 0\) such that \(f(B_{X}(x_{0}, \delta)) \subseteq B_{Y}(y_{0}, \varepsilon) \subseteq G\), that is \(B_{X}(x_{0}, \delta) \subseteq f^{-1}(G)\).
Theorem (Characterisations of continuity)
Let \(f : X \to Y\). The following are equivalent:
- \(f\) is continuous;
- \(f^{-1}(G)\) is open in \(X\) for every open \(G \subseteq Y\);
- \(f^{-1}(F)\) is closed in \(X\) for every closed \(F \subseteq Y\);
- \(f(\overline{B}) \subseteq \overline{f(B)}\) for every \(B \subseteq X\) (closure in \(X\) on the left, in \(Y\) on the right).
Proof: \((1) \implies (2)\): Proved in the previous lemma.
\((2) \implies (1)\): Given \(\varepsilon > 0\) and \(a \in X\), \(B_{Y}(f(a), \varepsilon)\) is open, so \(f^{-1}(B_{Y}(f(a), \varepsilon))\) is open. Since \(a \in f^{-1}(B_{Y}(f(a), \varepsilon))\), there exists \(\delta > 0\) with \(B_{X}(a, \delta) \subseteq f^{-1}(B_{Y}(f(a), \varepsilon))\), i.e. \(f(B_{X}(a, \delta)) \subseteq B_{Y}(f(a), \varepsilon)\).
\((2) \iff (3)\): If \(F \subseteq Y\) is closed, then \(Y \setminus F\) is open and \(f^{-1}(Y \setminus F) = X \setminus f^{-1}(F)\). Hence \(f^{-1}(F)\) is closed if and only if \(f^{-1}(Y\setminus F)\) is open.
\((3) \iff (4)\): \(f(\overline{B}) \subseteq \overline{f(B)}\) for every \(B\) is equivalent to \(\overline{B} \subseteq f^{-1}(\overline{f(B)})\) for every \(B\). Taking \(B = f^{-1}(F)\) with \(F\) closed, one obtains \(\overline{f^{-1}(F)} \subseteq f^{-1}(F)\), that is \(f^{-1}(F)\) closed, and conversely.
Theorem (Characterisation by sequences)
Let \(f : X \to Y\) and \(a \in X\). Then \(f\) is continuous at \(a\) if and only if for every sequence \(\{ x_{n}\}_{n \in \mathbb{N}} \subseteq X\) with \(x_{n} \to a\) we have \(f(x_{n}) \to f(a)\).
Proof: \((\implies)\): If \(x_{n} \to a\) and \(\varepsilon > 0\), by continuity there exists \(\delta > 0\) with \(d(x, a) < \delta \implies \rho(f(x), f(a)) < \varepsilon\). Take \(n_{0}\) such that \(n \geq n_{0}\) implies \(d(x_{n}, a) < \delta\); then \(\rho(f(x_{n}), f(a)) < \varepsilon\), that is \(f(x_{n}) \to f(a)\).
\((\impliedby)\): If \(f\) were not continuous at \(a\), there would exist \(\varepsilon > 0\) such that for every \(\delta > 0\) there exists \(x_{\delta} \in X\) with \(d(x_{\delta}, a) < \delta\) and \(\rho(f(x_{\delta}), f(a)) \geq \varepsilon\). Taking \(\delta = 1/n\) one obtains \(x_{n}\) with \(d(x_{n}, a) < 1/n\) and \(\rho(f(x_{n}), f(a)) \geq \varepsilon\). Thus \(x_{n} \to a\) but \(f(x_{n}) \not\to f(a)\), a contradiction.
Homeomorphisms and isometries
Definition (Homeomorphism)
Let \(f : X \to Y\). We say that \(f\) is a homeomorphism if \(f\) is continuous, bijective and \(f^{-1}\) is continuous. In that case, \(X\) and \(Y\) are said to be homeomorphic.
Note (Homeomorphisms map open sets to open sets)
If \(f : X \to Y\) is a homeomorphism and \(A \subseteq X\) is open, then \(f(A) = (f^{-1})^{-1}(A)\) is open in \(Y\).
Exercise (Homeomorphisms preserve the interior)
If \(f : X \to Y\) is a homeomorphism and \(A \subseteq X\), then \(f(A^{\circ}) = (f(A))^{\circ}\).
Definition (Isometry)
Let \(\phi : (X, d_X) \to (Y, d_Y)\). We say that \(\phi\) is an isometry if \(\phi\) is surjective and preserves the metric, that is, if for all \(x,y \in X\),
\[d_Y(\phi(x), \phi(y)) = d_X(x,y).\]Note (Every isometry is bijective)
If \(\phi : (X, d_X) \to (Y, d_Y)\) is an isometry, then \(\phi\) is injective, since if \(\phi(x) = \phi(y)\) then \(d_Y(\phi(x),\phi(y))=d_X(x,y) = 0\), whence we have that \(x=y\).