# Stone-Čech compactification of $\mathbb R$

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.

• People are already voting to close... It might help if you could say what you have tried? What's your definition of the Stone-Cech compactification? Do you have a start of a proof of the claim, but that you get stuck somewhere? What's your motivation? – Matthew Daws Mar 3 '12 at 21:29
• I think this is a good question. If $I$ is closed, then $\mathbb R\setminus I$ is homeomorphic to two copies of $\mathbb R$, so the answer is no (same for the half-closed case). If $I$ is open, it feels like the answer should be yes ... it would be necessary to show that $I$ remains open in $\beta\mathbb R$, which seems right, but Stone-Cech compactifications are weird and I don't see that right now. – Anton Geraschenko Mar 3 '12 at 21:46
• I should add that I haven't voted to close-- but the person who did hasn't given a reason... – Matthew Daws Mar 3 '12 at 22:17
• @Anton: If $I$ is closed then continuous (complex valued) functions on $\mathbb{R}\setminus I$ do not extend to continuous functions on $\mathbb{R}$. They do when $I$ is the open interval. – George Lowther Mar 4 '12 at 0:33
• @Mariarty: I think it would be better to ask your "update" as a new question... – Matthew Daws Mar 4 '12 at 14:54

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.)

• In some sense this is nothing but a careful elaboration of George Lowther's hint! – Matthew Daws Mar 4 '12 at 8:35
• So, why doesn't this argument also work for $U=X$? – George Lowther Mar 4 '12 at 9:44
• I think $U$ is implicitly assumed relatively compact: "... then $F\in C_0(X)$" – BS. Mar 4 '12 at 10:29
• I think $C(\beta X\setminus U)$ should be identified with $C_b(X)/C_0(U)$. – George Lowther Mar 4 '12 at 10:32
• @George, @BS: I've added the hypothesis that U is relatively compact. Thanks! George's 2nd comment is what's works for general $U$; and so I guess if $U$ isn't relatively compact, then $\beta X\setminus U$ is not $\beta(X\setminus U)$. – Matthew Daws Mar 4 '12 at 14:56

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.