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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.

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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
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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
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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
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@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
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@Mariarty: I think it would be better to ask your "update" as a new question... –  Matthew Daws Mar 4 '12 at 14:54
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2 Answers 2

up vote 6 down vote accepted

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.)

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In some sense this is nothing but a careful elaboration of George Lowther's hint! –  Matthew Daws Mar 4 '12 at 8:35
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So, why doesn't this argument also work for $U=X$? –  George Lowther Mar 4 '12 at 9:44
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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
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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.

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