Let $X$ and $Y$ be metric spaces and $E$ be a dense subset of X. Show that if $f\mathop: X \to Y$ and $g\mathop: X \to Y$ are continuous and $f(x) = g(x)$ for all $x \in E$, then $f(x) = g(x)$ for all $x \in X$.

Greetings, I am trying to prove this using the $\epsilon -\delta$ characterization of continuity. I already know how to prove it with the sequential characterization. Here is what I have so far:

Proof. Let $E$ be a dense subset of $X$. Let $f:X \to Y$ and $g: X\to Y$ be continuous on $X$ such that $f(x) = g(x)$ for all $x\in E$. Suppose to the contrary that $f(x) \neq g(x)$ for some $x\in X$. Since $E$ is dense in $X$, $\overline{E} = X$. Hence, there exists $x_0 \in E'$ such that $f(x_0) \neq g(x_0)$. Let $\epsilon > 0$. Choose $\delta_1>0$ such that $x\in X$ and $d_X(x,x_0)<\delta_1$ implies $d_Y(f(x),f(x_0)) < \frac{\epsilon}{2}$. Choose $\delta_2 > 0$ such that $x\in X$ and $d_X(x,x_0)< \delta_2$ implies $d_Y(g(x),g(x_0)) < \frac{\epsilon}{2}$. Put $\delta = \min\{\delta_1,\delta_2\}$. Then if $x\in E$ and $d_X(x,x_0)< \delta$ we have $$d_Y(f(x_0),g(x_0)) \leq d_Y(f(x_0),f(x)) + d_Y(f(x),g(x_0)) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ which implies that $f(x_0) = g(x_0)$, a contradiction.

Now, is this sufficient to prove the statement? Notice that my inequality is only true for $x\in E$ since the assumption is that $f(x) = g(x)$ for all $x\in E$. I think I need to consider $x\in E'$ and show a similar inequality. Thanks for your help.


Just change "Then if $x \in E$ and $d_X(x,x_0) < \delta...$" by "There exists $x \in E$ such that $d_X(x,x_0) < \delta$..." and the proof follows. You can change because $x_0$ is in the closure and then for every open set around $x_0$ exists $x \in E$ such that $d_X(x,x_0) < \delta$.

  • $\begingroup$ Great point. Thanks. $\endgroup$ – johnny133253 Nov 3 '18 at 6:08

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