One Point Compactification and Stone-Čech compactification question

Question

'Let $X$ Hausdorff locally compact space such that every continuous map $f: X \to \mathbb R$ can be extended to a continuous map $g: X^{\ast} \to \mathbb R$.

Prove that $X^{\ast} = \beta (X)$.

I have two ideas:

i) $X$ Hausdorff locally compact, then $X$ is completely regular and this is eqivalent to $X$ is homeomorphic to a subspace of $\displaystyle \prod_{h\in H} [0,1]$ with

$H$ = { $h:X\to [0,1]: h\ \text{continuous}$}, and $X^{\ast}$ is Hausdorff compact, then normal, and normal implies completely regular (and then is homeomorphic to a subspace of $\displaystyle \prod_{h\in H^{\ast}} [0,1]$ with $H^{\ast}$ = { $h:X^{\ast}\to [0,1]: h\ \text{continuous}$}). If I could prove that $|H| = |H^{\ast}|$, then I can finish the proof.

ii) Let $i: X\to \beta(X)$ and $f:X \to K$ continuous with $K$ Hausdorff compact space, then exists only one continuous function $g: \beta(X) \to K$ such that $g\circ i = f$ and $\beta(X)$ is the unique compactification that meets this property. So, if i can replace $\beta(X)$ with $X^{\ast}$ and $X^{\ast}$ meets the previous property, then $\beta(X) = X^{\ast}$

I don't know if my ideas are the best, but in both cases i'm not sure how to use the hypotesis of the continuous extension of the functions $f:X\to\mathbb R$

Edit: Let $i:X \to \displaystyle \prod_{h\in H} [0,1]$ with $H$ = { $h:X\to [0,1]: h\ \text{continuous}$} defined by $(i(x))_h = h(x)$, then $\overline{i(X)} = \beta(X)$ is the Stone-Čech compactification

Answer 1

Recall the universal property of $\beta X$:

If $K$ is a compact Hausdorff space, then every continuous function $X\to K$ has a continuous extension on $\beta X$.

In fact, $\beta X$ (constructed by embedding $X$ in a product of unit intervals) is the unique Hausdorff compactification of $X$ with such property, up to homeomorphism. Indeed, if $Y$ is another such space, then let $\epsilon\colon X\to\beta X$ and $\eta\colon X\to Y$ be the embeddings. Then they have continuous extensions $\epsilon_*\colon Y\to\beta X$ and $\eta_*\colon\beta X\to Y$ respectively. Now $\eta_*\circ\epsilon_*$ coincides with $i_Y$ on $X$, viewed as a dense subset of $Y$. Since both are continuous, they agree on $Y$. Similarly, $\epsilon_*\circ\eta_*=i_{\beta X}$. We conclude that $\epsilon_*$ and $\eta_*$ are the inverses of each other, so $\beta X$ is homeomorphic to $Y$.

We use this universal property to prove the following theorem:

If $Y$ is a Hausdorff compactification of $X$, and if every bounded continuous real function on $X$ has a continuous extension on $Y$, then $Y$ is homeomorphic to $\beta X$.

Let $K$ be an arbitrary compact Hausdorff space, and $I$ be the set of all continuous functions $K\to[0,1]$. Embed $K$ in $[0,1]^I$ through $\epsilon$. Given continuous $f\colon X\to K$, for every $g\in I$ the function $g\circ f$ has a continuous extension $g_*\colon Y\to [0,1]$. The function $G\colon Y\to[0,1]^I$ by $$ G(x)_g=g_*(x) $$ is a continuous extension of $\epsilon\circ f$. Hence every continuous function $X\to K$ has a continuous extension on $Y$.

In your case, every bounded real function on $X$ has a continuous extension from $X^*$ into $\mathbb{R}^*$. But since the function is bounded, so is its extension. Hence $X^*$ is (homeomorphic to) $\beta X$.

Answer 2

A simpler argument goes as follows. You have $X$, you have the canonical map $\phi:X\to \prod_{h\in H}[0,1]$ given by $\pi_h\phi(x)=h(x)$, and you have that $\beta X$ is the closure of $\phi(X)$ within $\prod_{h\in H}[0,1]$. Now, you can prove that $\phi$ extends to a map $\tilde\phi:X^\ast\to \prod_{h\in H}[0,1]$. By the universal property of the product, you just need to define each coordinate $\pi_h\tilde\phi:X^\ast\to [0,1]\subset\mathbb R$, and this can be taken as the extension of $h$ given by the hypothesis. Then, $\tilde\phi(X^\ast)=\phi(X)\cup\{\tilde\phi(\infty)\}$ is compact, therefore closed, and you can conclude that $\beta X=\tilde\phi(X^\ast)\cong X^\ast$.