Dense subset maps under continous map to whole of its range.

Question

Let $f,g$ be continous maps on $X$.Let $E$ be dense subset of $X$.Then $\overline{f(E)}=f(X)$ and if $f(p)=g(p) \forall p\in E$ then $f(p)=g(p) \forall p\in X$.

Part 1: Show that $\overline{f(E)}=f(X)$
As $E$ is dense subset $\overline{E}=X$ So we have to prove that $\overline{f(E)}=f(\overline{E})$

Claim 1:$f(\overline{E})\subset \overline{f(E)}$
$\overline{E}=E\cup E'$
$f(E\cup E')=f(E)\cup f(E')$
So $f(E)\subset \overline{f(E)}$
$\overline{f(E)}=f(E)\cup f(E)'$
Only thing remains to show that $f(E')\subset f(E)'$

Take $x\in E'$ so $\forall \epsilon > \exists y\in E $ such that $d(x,y)<\epsilon$
By continuity of f on X , so f is continous at x $\forall \epsilon > 0,\exists \delta >0 ,y\in E , s.t. d(x,y)<\delta \to d(f(x),f(y))<\epsilon $ SO this is true for every $\epsilon $ so f(x) is limit point of $f(E)$ so $f(x)\in f(E)'$

So $f(\overline{E})\subset \overline{f(E)}$

Claim 2:$ \overline{f(E)}\subset f(\bar{E})$. i.e $f(E)'\subset f(\overline{E})=f(X) $
as $X=\overline{E}$

$ y\in f(E)'$ then we have to show that $\exists z\in X $ such that $f(z)=y $

$ y\in f(E)'$ then $\forall r>0 \exists f(x)\in f(E)$ such that $d(f(x),y)<r$
I don't able to use continutiy assumption with above to show that $\exists z \in X$ such that f(z)=y

Part 2: if $f(p)=g(p) \forall p\in E$ then $f(p)=g(p) \forall p\in X$.

On contary suppose $\exists y\in E^c$ such that $f(y)\neq g(y)$ SO $\exists r>0$ such that $d(f(y),g(y))>r$
By continity of both f and g on X for $y\in X$ and denseness of E
$\forall r>0 , \exists q\in E$ such that $d(y,q)<r\to d(f(y),f(q))<r/3 ,d(g(y),g(q))<r/3$
A $q\in E$ then $f(q)=g(q) $ ,i.e $d(f(q),g(q))<r/3$
So $\forall r>0$ we have $d(f(y),g(y))<d(f(y),f(q))+d(g(y),g(q))+d(f(q),g(q))<r$ which is contradicts with given .

Hence DOne

Any Help will be appreciated

Answer 1

The result is not correct as you have stated it. For example, a continuous inclusion of an open set into a topological space. Think about $f:(0,1)\rightarrow [0,1]$ defined by $f(x) = x$. Take $E = (0,1)$. Then $\overline{E} = (0,1)$ since $E$ is already the entire space $(0,1)$, so it is both open and closed. But $(0,1) = f(\overline{E})\neq \overline{f(E)} = [0,1]$

What you mean to prove is that $f(E)$ is dense in $f(\overline{E})$. That means that for every closed set $C$ with $f(E)\subset C$, we also have $f(\overline{E})\subset C$. So let $C$ be some closed set containing $f(E)$. $f^{-1}(C)$ is a closed set, since $f$ is continuous and $C$ is closed. $f(E)\subset C$ implies that $E\subset f^{-1}(C)$. But $f^{-1}(C)$ is closed, so it also contains the closure $\overline{E}$. Then $\overline{E}\subset f^{-1}(C)$ implies that $f(\overline{E})\subset C$, which is what we wanted to show.

Answer 2

From a topological viewpoint very similar counterexample to the one by user585541, but more "calculus-like". Put $E=X=\mathbb R$ and consider $f(x)=\arctan(x)$. Then $\overline{ f(E)} =\overline{f(\mathbb R)}=\overline{(-\pi/2,\pi/2)}=[-\pi/2,\pi/2] \neq(\pi/2,\pi/2)=f(\mathbb R)=f(X)$.

The second part is correct: two continuous functions $f,g$ coinciding on a dense subset $E$ of their (identical) domain $X$, coincide on the whole domain $X$. Again, more "calculus-like" approach is to realize that any point $x \in X$ is a limit of some sequence of points $x_n \in E$, $n=1,2,\dots$ (because $E$ is dense in $X$) and then use the continuity of the functions: $f(x)=\lim f(x_n)=\lim g(x_n)=g(x).$