# Embeding a space into the product space

Let $X$ be a Hausdorff space. Let $\mathscr{F}=\left\{f_j:X\rightarrow R\mid j\in J\right\}$ be a family of continuous real-valued functions with following property:for every $x\in X$, and every closed set $A\subset X$ with $x\notin A$, there exists $f_j\in \mathscr{F}$ with $f_j (x)>0$ and $f_j(a)=0$ for every $a\in A$. Define $F:X\rightarrow R^J$ via $F(x)=(f_j(x))_{j\in J}$. Show that $F$ is an embedding of $X$ into $R^J$.

First of all, (injection) in Hausdorff space,every single point is closed. So if $F(x_1)=F(x_2)$ then $x_1=x_2$ since $f_j$'s could separate the points in $X$.

Secondly,since each $f_j$ is continuous, so does $F(x)$ from the property of product topology.

Then I'm stuck with the proof of inverse of $F(x)$ is continuous. Could anyone give me some help?

HINT: Equivalently, you can show that $F$ is an open map. Suppose that $U$ is an open set in $X$. For each $x\in U$ there is a $\varphi(x)\in J$ such that $f_{\varphi(x)}(x)>0$, and $f_{\varphi(x)}(y)=0$ for each $y\in X\setminus U$. Now suppose that $x\in U$; we want to find an open nbhd $V$ of $F(x)=\langle f_j(x):j\in J\rangle$ in $\Bbb R^J$ such that $V\cap F[X]\subseteq F[U]$. In fact it’s possible to choose $V$ to be a subbasic open nbhd of $F(x)$, i.e., one that restricts just one coordinate to a proper open subset of $\Bbb R$, and the setup that I gave you should point you in the right direction to find that coordinate.
• As your hint pointed, the $V$ should be chosen as the open subbasic open set with $\varphi(x)$th coordinate $(0,\infty)$ the rest coordinates $R$. But how can we guarantee that $V\cap F[X]$ belongs to $F[U]$? – mike Nov 27 '16 at 2:54
• yes,the right set is $R-{0}$,and since $V\cap F[X]$ doesn't belong to $F[X/U]$ from the fact that $f_{\varphi(x)}(y)=0$ for each $y\in X/U$.So the inclusion comes out. Thank you for your hint. – mike Nov 27 '16 at 3:38
• @mike: Oops! Sorry, I was thinking about what happens back in $X$. Of course you were absolutely right, and restricting to the strictly positive reals works fine even if the function also takes negative values: $\{y\in X:f_{\varphi(x)}(y)>0\}$ is certainly a subset of $U$. But $\Bbb R\setminus\{0\}$ also works. You’re welcome, and I apologize for the misleading comment, which I’m now deleting. – Brian M. Scott Nov 27 '16 at 3:43