Continuous Functions are Determined on Dense Sets

Last updated Nov 1, 2022

Suppose $(X, \tau)$, $(Y, \rho)$ are Topological Spaces, $Y$ is Hausdorff, and $f: X \to Y$, $g: X \to Y$ are continuous functions. Suppose $D \subset X$ is Dense in $X$. If $$g {\big|}{D} = f {\big|}{D}$$ we have that $g = f$.

First observe that since $D$ is Dense its Closure is $X$. Because Closure of a Set is all its Net Limits, so for every $x \in X$, there exists a Net $(x_{\alpha}){\alpha \in A} \subset D$ so that $x{\alpha} \to x$. Recall that A Function is Continuous iff it preserves Net Convergence, so $f(x_{\alpha}) \to f(x)$ and $g(x_{\alpha}) \to g(x)$. Net Limits are unique in Hausdorff Spaces so, because $f(x_{\alpha}) = g(x_{\alpha})$ for all $\alpha \in A$, we must have that $f(x) = g(x)$. $x \in X$ was arbitrary so $f = g$. $\blacksquare$

1. This is not true in general. Suppose $X={1, 2, 3}$ is equipped with the Indiscrete Topology. Suppose $f = \text{id}_{X}$ and $g: X \to X$ s.t. $$g(x) = \begin{cases} 1 & \text{if } x = 1\\ 3 & \text{if } x = 2\\ 2 & \text{if } x = 3\\ \end{cases}$$ Then $f$ is Continuous Function since Identity Functions are Continuous. $g$ is continuous since All Surjective Functions to the Indiscrete Topology are Continous. All Nonempty Subsets of the Indiscrete Topology are Dense, so ${1}$ is Dense. We see $g(1) = f(1)$, but $g(2) = 3 \neq 2 = f(2)$. $\blacksquare$
2. Since Metric Spaces are First Countable and Hausdorff, continuous functions between Metric Spaces are determined on a dense set.