# Complex problem in Hausdorff

#### bw0young0math

##### New member
Hello. My friend asked me thos problem today, but I couldn't.
Here is the problem.

X: topological space&T2(Hausdorff)
D: dense subet of X
f:X→Y is continuous function and
restriction fuction of f to D i.e., f(D) is embedding fuction.

Show that f(X-D)<Y-f(D) (<means set inclusion. i.r., Y -f(D) includes f(X-D). )

I wanted to solve it as using Reduction absurdity.
Thus I assumed that there exists a y in f(X-D) but not in Y-f(D).
Therefore I supposed that f(X-D) < f(D).

then.. how can I solve it with many conditions above?

#### Opalg

##### MHB Oldtimer
Staff member
Re: complex problem in Hausedorff.

Hello. My friend asked me thos problem today, but I couldn't.
Here is the problem.

X: topological space&T2(Hausdorff)
D: dense subet of X
f:X→Y is continuous function and
restriction fuction of f to D i.e., f(D) is embedding fuction.

Show that f(X-D)<Y-f(D) (<means set inclusion. i.r., Y -f(D) includes f(X-D). )

I wanted to solve it as using Reduction absurdity.
Thus I assumed that there exists a y in f(X-D) but not in Y-f(D).
Therefore I supposed that f(X-D) < f(D).

then.. how can I solve it with many conditions above?

I found it easiest to do this problem first in the case where $X$ is a metric space, so that we can use convergent sequences to detect closure and continuity.

So let $x\in X$ with $f(x)\in f(D)$, say $f(x) = f(y)$ for some $y$ in $D$. Since $D$ is dense on $X$ there is a sequence $(x_n)$ in $D$ with $x_n\to x$. By continuity, $f(x_n) \to f(x) = f(y)$. But $f: D\to f(D)$ is an embedding, which means that it is a homeomorphism and so the inverse map from $f(D)$ to $D$ is continuous. Thus $f(x_n)\to f(y)$ implies that $x_n\to y$. But we already know that $x_n\to x$, and so $x=y\in D$. This shows that if $f(x)\in D$ then $x\in D$, which implies that if $x\in X-D$ then $f(x)\in Y-f(D)$.

Having seen how to do it in the metric case, you now need to find an analogous argument in the general case. As before, let $x\in X$ with $f(x)\in f(D)$, say $f(x) = f(y)$ for some $y$ in $D$. Suppose that $x\ne y$, then by the Hausdorff condition there exist disjoint open neighbourhoods $U$ of $x$ and $V$ of $y$. The map $f^{-1}:f(D)\to D$ is continuous, so there exists an open neighbourhood $W$ of $f(y)$ such that if $z\in D$ and $f(z)\in W$ then $z\in V$. But $f$ is continuous, and $f(x)\in W$, so there exists an open neighbourhood (I'm running out of suitable letters) $T$ of $x$ such that $f(T)\subset W$. If $z\in T\cap U\cap D$ then $f(z)\in W$ and so $z\in V$. But $z$ cannot be in $U$ and $V$, so no such $z$ can exist. Thus $T\cap U$ is a neighbourhood of $x$ containing no elements of $D$. In other words, $x$ is not in the closure of $D$. But that contradicts the density of $D$. The conclusion is that $x=y$ and, as in the metric case, that wraps up the proof.

#### bw0young0math

##### New member
Re: complex problem in Hausedorff.

I
I found it easiest to do this problem first in the case where $X$ is a metric space, so that we can use convergent sequences to detect closure and continuity.

So let $x\in X$ with $f(x)\in f(D)$, say $f(x) = f(y)$ for some $y$ in $D$. Since $D$ is dense on $X$ there is a sequence $(x_n)$ in $D$ with $x_n\to x$. By continuity, $f(x_n) \to f(x) = f(y)$. But $f: D\to f(D)$ is an embedding, which means that it is a homeomorphism and so the inverse map from $f(D)$ to $D$ is continuous. Thus $f(x_n)\to f(y)$ implies that $x_n\to y$. But we already know that $x_n\to x$, and so $x=y\in D$. This shows that if $f(x)\in D$ then $x\in D$, which implies that if $x\in X-D$ then $f(x)\in Y-f(D)$.

Having seen how to do it in the metric case, you now need to find an analogous argument in the general case. As before, let $x\in X$ with $f(x)\in f(D)$, say $f(x) = f(y)$ for some $y$ in $D$. Suppose that $x\ne y$, then by the Hausdorff condition there exist disjoint open neighbourhoods $U$ of $x$ and $V$ of $y$. The map $f^{-1}:f(D)\to D$ is continuous, so there exists an open neighbourhood $W$ of $f(y)$ such that if $z\in D$ and $f(z)\in W$ then $z\in V$. But $f$ is continuous, and $f(x)\in W$, so there exists an open neighbourhood (I'm running out of suitable letters) $T$ of $x$ such that $f(T)\subset W$. If $z\in T\cap U\cap D$ then $f(z)\in W$ and so $z\in V$. But $z$ cannot be in $U$ and $V$, so no such $z$ can exist. Thus $T\cap U$ is a neighbourhood of $x$ containing no elements of $D$. In other words, $x$ is not in the closure of $D$. But that contradicts the density of $D$. The conclusion is that $x=y$ and, as in the metric case, that wraps up the proof.
Thanks to explaining with solution in metric space, it was not difficult understanding problem. Thank you.

I wrote solution on my way with refering to your solution.
^^* : ³×ÀÌ¹ö ºí·Î±×

(Of course, I want to write my solution here but it is difficult to me. Sorry.)

#### MarkFL

Personally, I find the images in the link hard to read. There isn't quite enough resolution in the images for these old eyes. I'm not trying to be critical, but honest. I highly recommend for posting such proofs that you look into the use of $\LaTeX$. There are many online tutorials, and its use is fairly straightforward. The end result is something that is much easier for our members to read. (flower)