Question

Let $\beta X$ - is a Stone-Čech compactification of $X$. $I=(-1,1)$ - is an interval of the real line. Is it true that $\beta \mathbb R\setminus I = \beta(\mathbb R\setminus I)$? In other words, it means that a finite interval does not affect on the "compactification of infinity".

Update: Great thanks for realized calculations.

Answer 1

I can show the following (which Anton was asking about in comments). Let $X$ be locally compact and Hausdorff, and $U\subseteq X$ open. Let $X_\infty$ be the one-point compactication, so $U$ is still open in $X_\infty$. By the universal property of the Stone-Cech compactification, there is a continuous map $\phi:\beta X\rightarrow X_\infty$ which is the identity on $X$. Then $\phi^{-1}(U)$ is open in $\beta X$, and is just the canonical image of $U$ in $\beta X$. So $U$ open in $X$ shows that $U$ is open in $\beta X$.

(This fails for general closed sets. If $F\subseteq X$ is closed, then $F$ is only closed in $X_\infty$ if $F$ is also compact.)

I'll now use that $\beta X$ is the character space of $C^b(X)$. Let $U\subseteq X$ be open.

Lemma: Assume that $U$ is relatively compact. Under the isomorphism $C(\beta X)=C^b(X)$, we identify the ideal $\{ f\in C(\beta X) : f(x)=0 \ (x\not\in U) \}$ with $\{ F\in C^b(X) : f(x)=0 \ (x\not\in U) \}$

Proof: $X$ is itself open in $\beta X$, and the image of $C_0(X)$ in $C(\beta X)$ is just the functions vanishing off $X$. If $F\in C^b(X)$ vanishes off $U$ then $F\in C_0(X)$ (as $U$ is relatively compact) and so the associated $f$ in $C(\beta X)$ vanishes off $U$. Conversely, if $f\in C(\beta X)$ vanishes off $U$ then the associated $F\in C^b(X)$ is just the restriction of $f$ to $X$, and so vanishes off $U$.

By the Tietze theorem, the restriction map $C(\beta X) \rightarrow C(\beta X \setminus U)$ is a surjection. So we can identify $C(\beta X\setminus U)$ with the quotient $C(\beta X) / \{ f\in C(\beta X) : f(x)=0 \ (x\not\in U) \}$. So by the above, we identify $C(\beta X \setminus U)$ with $C^b(X) / \{ F\in C^b(X) : F(x)=0 \ (x\not\in U) \}$. If $X$ is normal, then we can again use Tietze to extend any $F\in C^b(X\setminus U)$ to all of $X$. It follows that $C^b(X) / \{ F\in C^b(X) : F(x)=0 \ (x\not\in U) \}$ is isomorphic to $C^b(X\setminus U) = C(\beta(X\setminus U))$. So $\beta X \setminus U = \beta (X\setminus U)$ (in a fairly canonical way) under the hypotheses that $X$ is normal and $U$ is relatively compact.

(I'm not sure what happens for non-normal $X$. For $X=\mathbb R$ and $U$ an open interval, we obviously don't need Tietze.)

Answer 2

More generally: if $X$ is normal and $A$ is closed in $X$ then, by the Tietze-Urysohn theorem, the closure in $\beta X$ of $A$ is $\beta A$. In the example above $X=\mathbb{R}$ and $A=\mathbb{R} \setminus (-1,1)$. As the closure of $(-1,1)$ in $\beta\mathbb{R}$ is just $[-1,1]$ the desired equality follows.