I am learning Topology from Dolciaini Expository text in Topology by S.G Krantz.
I have question in proof of following theorem:
Here $I^{C}$ denote the collection of all functions from C to I.
I am not able to prove $f$ to be embedding.
For proving 1. $f$ is 1-1 .
$f$ is continuous.
$f^{-1}$ is continuous on image of $f$.
$f(x)$ is clearly continuous. let $f(x)=f(y)$ , then if $f^{-1}$ exists we can prove it 1-1 . But I also can't deduce on why $f^{-1} $ exists. I don't have any other thoughts except these.
Can you please tell how to prove 1 and 3 in defination of embedding.
It is a good idea to set up proper notations for the map. For $x \in X$ write $F_x$ for the map defined on $C$ by $F_x(f)=f(x)$. The the map under consideration is $x \to F_x$.
If the image of $x$ is same as image of $y$ then $F_x=F_y$ so $F_x(f)=F_y(f)$ of $f(x)=f(y)$ for all $f \in C$. By complete regularity $x \neq y$ implies there exists $f\in C$ such that $f(x)=1$ and $f(y)=0$. This contradiction shows that $x=y$. Hence our map is one-to-one.
Hint for continuity of the inverse: Let $U$ be an open set in $X$ . It is enough to show that the image of $U$ is an open subset of the range of our map. Let $x \in U$ There exists $f \in C$ such that $f(x)=1$ and $f(y)=0$ for all $y \in X\setminus U$. Consider $\{F_t: F_t(f) >\frac1 2\}$. Verify that this is open, contains $F_x$ and it is contained in the image of $U$.