Stone Cech compactification of $(0,1]$is not $[0,1]$

Question

If I considered $(0,1]\subset \beta(0,1]=[0,1]$ this is trivial because the universal property would mean we would need a continuos function on $[0,1]$ with $g(t)=\sin(1/t)$ fot all $t \in(0,1]$ which is clearly absurd (no $g(0)$ makes $g$ continuous). I do not want to use this continence, because I only assume $\beta(0,1]\approx [0,1]$ and so, $(0,1]$ could be "scrambled inside" of $[0,1]$.

Here is what I tried so far:

I am interested in proving the Stone Cech compactification of $(0,1]$ is not homeomorphic to $[0,1]$.

Because $[-1,1]$ is a Hausdorff compact space, the universal property of the Stone Cech compactification teaches us there would be a unique continuos map $g:\beta (0,1]\rightarrow[-1,1] $ such that $g(\epsilon(t))=\sin(1/t)$ where $\epsilon$ is the evaluation function given by $\epsilon(t)=(f(t))_{f\in C_b(0,1]}$.

Let $\Phi: \beta(0,1]\rightarrow [0,1] $ be a homeomorphism. In this case, $g\Phi^{-1}\Phi\epsilon(t)=\sin(1/t)$ for $1\geq t>0$. My initial tought was taking $t=\frac{1}{n}$ in this case $\Phi\epsilon(1/n)\in [0,1]$ is bounded and by Bolzano Weierstrass there is a convergent subsequence, such that $\Phi\epsilon(1/n_k)\rightarrow x_o$. So, we have $\sin(n_k)=\sin(1/(1/n_k))\rightarrow g\Phi^{-1}(x_o)$. I thought this would be a contradiction but it appears there are always (Bolzano Weierstrass again) convergent subsequences of the form $\sin(n_k)$, so it could be $\sin(n_k)$ indeed converges.

Answer 1

The Stone Cech compactification is the biggest compactification of a completely regular space, the one-point compactification is the smallest compactification of a locally compact space.

Clearly $[0,1]$ is the one-point compactification of $(0,1]$. If it also were the Stone Cech compactification of $(0,1]$, then each map $f : (0,1] \to K$ to a compact Hausdorff space $K$ would have a (unique) continuous extension $\bar f : [0,1] \to K$.

The space $(0,1]$ has many compactifications, for example the closed topologist's sine-curve $T$. See Non-equivalent compactifications of $[0, \infty)$. The identity on $(0,1]$ should therefore have a continuous extension $\phi = \bar id : [0,1] \to T$. Since $(0,1]$ is dense in $T$, we have $\phi([0,1]) = T$ which implies that $\phi(\{0\})$ is the remainder $R = T \setminus(0,1]$. But this is impossible because $R$ has more than one point.

Update:

$\phi(\{0\}) = R$ is a consequence of the following general

Lemma. Let $f : X \to Y$ be a continuous map from a compact space $X$ to a Hausdorff space $Y$ such that $f(X)$ contains a dense subspace $B \subset Y$. Then $f(X) = Y$. In particular the remainder $R = Y \setminus B$ is contained in $f(X)$.

Proof. $f(X)$ is a compact subset of $Y$, thus it is closed in $Y$. We have $Y = \overline B \subset \overline{f(X)} = f(X)$, hence $f(X) = Y$. $\quad \square$

Now consider $\phi : [0,1] \to T$. It maps $(0,1]$ bijectively onto the dense subspace $(0,1]$ of $T$. The Lemma shows that the remainder $R$ must be contained in $\phi([0,1])$. Thus $\phi(\{0\}) = R$.

Update:

Let us be a bit more formal.

The Stone Cech compactification is a dense embedding $i : (0,1] \to \beta(0,1]$ into a compact Hausdorff space $\beta(0,1]$ satisfying the well-known universal property. Often one says that the space $\beta(0,1]$ is the Stone Cech compactification, but the embedding $i$ is an essential part of the construct.

Assume that there exists a homeomorphism $h : \beta(0,1] \to [0,1]$. Then $\epsilon = h \circ i : (0,1] \to [0,1]$ is a dense embedding which also has the universal property. Using this $\epsilon$ we could show that it is possible to replace $h$ by a homeomorphism $h' : \beta(0,1] \to [0,1]$ such that $\epsilon' = h' \circ i : (0,1] \to [0,1]$ is the inclusion $\iota : (0,1] \hookrightarrow [0,1]$ which would imply that $\iota$ must have the universal property. But we shall not need this.

Since $\epsilon((0,1])$ is connected, it must be a subinterval of $[0,1]$. The only two dense subintervals which are homeomorphic to $(0,1]$ are $(0,1]$ and $[0,1)$. The remainder $R_\epsilon = [0,1] \setminus \epsilon((0,1])$ is therefore a set with one element.

By the universal property the embedding $f : (0,1] \to T, f(t) = (t, \sin(1/t))$ admits a continuous $\bar f : [0,1] \to T$ such that $\bar f \circ \epsilon = f$. The Lemma implies that that the remainder $R = T \setminus f((0,1])$ is in the image of $\bar f$. Therefore we must have $R = \bar f(R_\epsilon)$. This is a contradiction because $R$ is an infinite set.