ThmDex – An index of mathematical definitions, results, and conjectures.
Result R80 on D47: Lipschitz map
Lipschitz map is continuous
Formulation 0
Let $X$ and $Y$ each form a D2506: Topological metric space.
Let $d_X$ and $d_Y$ each be the D58: Metric in $X$ and $Y$, respectively.
Let $f : X \to Y$ be a D47: Lipschitz map with respect to $X$ and $Y$.
Then $f$ is a D55: Continuous map with respect to $X$ and $Y$.
Proofs
Proof 0
Let $X$ and $Y$ each form a D2506: Topological metric space.
Let $d_X$ and $d_Y$ each be the D58: Metric in $X$ and $Y$, respectively.
Let $f : X \to Y$ be a D47: Lipschitz map with respect to $X$ and $Y$.
Denote by $M$ a constant for which $f$ satisfies the Lipschitz condition. We may assume that $M > 0$, since otherwise we could simply trade $M$ for any strictly greater constant. Let now $\varepsilon > 0$ and define $\delta = \varepsilon / M$. If $x, y \in X$ such that $d(x, y) < \delta$, then \begin{equation} d_Y(f(x), f(y)) \leq M d(x, y) < M \delta = M \frac{\varepsilon}{M} = \varepsilon \end{equation} Since $\varepsilon > 0$ was arbitrary, the claim follows from R703: Continuity in metric-topological space. $\square$