Let $S(X)=\prod_{U \in \tau_{X}} \{ 0,1 \}_{U} $ with $(X,\tau_{X})$ $T_0$ topological space and $\{ 0,1 \}$ the Sierpinski's space. Cosiderer the map $e$ from $X$ to $S(X)$. $e:X\rightarrow S(X)$ such that $e(x)_{U}=0$ if $x \in U$ and $e(x)_{U}=1$ if not. I've proved that $X \simeq e[X]$. Since $S(X)$ is compact hausdorff space, then $\beta_{2}(X)= \overline{e[X]}$ is a compactification for $X$. My question is:
How can I prove that $\beta_{2}(X)$ doesn't have the universal property?
I mean this property:
Given any continuous map $f:X\rightarrow C$ of $X$ into a compact Hausdorff space $C$, the map $f$ extends uniquely to a continuous map $g:\beta(X)\rightarrow C)$ .
$X$ is $T_{0}$ only.
I have this idea: to prove: For any $f$ real-valued function on $X$,$f :X\rightarrow \mathbb{R}$ we'll have the map $f$ don't extends uniquelyto a continuous map $g :\beta_{2}(X)\rightarrow \mathbb{R}$.
Help!