0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $d_X(x,x_0) < \delta$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.