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Let $X$ be a completely regular space and let $T$ a topological space such that $X \subseteq T \subseteq \beta X$. Then $\beta T = \beta X$, where $\beta$ denotes the Stone-Cech compactification.

Solution: Let $f: T \mapsto [0,1]$. Then it suffices to show that the restriction of f to $X$ can be extended to $\beta X$"..

Why is this?

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$X$ doesn't seem to be used here. Couldn't you just say you have $C\subseteq D\subseteq Z$? – Jonas Meyer Jan 17 '11 at 5:29
It doesn't. Are you sure there are no additional conditions on $X$? (say, is it Hausdorff? Regular?) or on $f$? or on $C$ and $D$? (say, is $C$ dense in $D$?) – Andrés Caicedo Jan 17 '11 at 5:29
I guess if $X$ is normal then we could use the following: – PEV Jan 17 '11 at 5:32
@Andres Caicedo: Just corrected it. The original post was a mess, I am sorry. – student Jan 17 '11 at 5:38

Here is my answer to the modified question: Because $X$ is dense in $\beta X$, it is also dense in $T$, and therefore every continuous function on $T$ is uniquely determined by its restriction to $X$.

Here is my answer to the original question, which asked when, in order to see that a continuous function $f:D\to [0,1]$ has a continuous extension to $Z$, it suffices to show that the restriction of $f$ to $C$ has a continuous extension to $Z$, where $C\subseteq D\subseteq Z$ and $Z$ is a topological space.:

This would suffice precisely when every continuous function from $D$ to $[0,1]$ is determined by its restriction to $C$. This will always be true if $C$ is dense in $D$. If $D$ is completely regular, then density of $C$ is also necessary. To see this, suppose $D$ is completely regular and $C$ is not dense. Then there is a continuous function $f:D\to [0,1]$ such that $f$ is zero on $\overline C$, but $f(x)=1$ for some $x\in D\setminus\overline C$. Since the zero function on $D$ is also a continuous extension of the zero function on $C$, the zero function on $C$ has more than one continuous extension to $D$.

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Thank you for your reply. – student Jan 17 '11 at 6:05
@student: You're welcome. And you can see the following if you want proof that a continuous function is determined by its restriction to a dense subset:… – Jonas Meyer Jan 17 '11 at 6:42

Recall that $\beta T$ is characterized by being (up to homeomorphism) the unique compact (Hausdorff) space of which $T$ is a dense subset and such that any continuous map $f:T\to[0,1]$ can be extended to a continuous map $g:\beta T\to[0,1]$.

Ok, so, given $f:T\to[0,1]$, continuous, we want to show that it extends to a continuous $g:\beta X\to[0,1]$.

First, $\beta T$ makes sense since $T$ is completely regular, being a subspace of $\beta X$.

Second, $X$ is dense in $T$, so any continuous $f:T\to[0,1]$ is completely determined by its restriction to $X$.

Third, by definition of $\beta X$, this restriction extends to a continuous $g:\beta X\to[0,1]$.

Since $f\upharpoonright X$ uniquely determined $f$, we have shown that any continuous $f:T\to[0,1]$ extends to a continuous $g:\beta X\to[0,1]$.

This proves the theorem: Since $X\subseteq T$, then $\beta X\subseteq \beta T$. If $\beta T$ happened to be larger than $\beta X$, then we have that any continuous $h:X\to[0,1]$ can be extended to a continuous $j:\beta T\to[0,1]$, contradicting that $\beta X$ is largest with this property. The reason why this extension exists, is because if $i:\beta X\to[0,1]$ is the continuous extension of $h$, then its restriction to $T$ is continuous, and can therefore be extended.

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@Andres Caicedo: Thank you. I know I'm blind. How do you show rigorously that $X$ is dense in $T$ ? – student Jan 17 '11 at 6:05
@student: $X$ is dense in $\beta X$, and $T$ sits in the middle. – Andrés Caicedo Jan 17 '11 at 6:06
@Andres Caicedo: If $X$ is dense in $\beta X$ then $T$ is dense in $\beta X$. This also implies $X$ is dense in $T$? – student Jan 17 '11 at 6:10
@student: If $A\subseteq B\subseteq C$ and $A$ is dense in $C$, then $A$ is dense in $B$. If $x$ is in $B$ and $U$ is a neighborhood of $x$ in $B$, then by the definition of the subspace topology, there is an open subset $V$ of $C$ such that $U=B\cap V$. Since $U\cap A=V\cap A$ and $A$ is dense in $C$, $U\cap A$ is nonempty, so $A$ is dense in $B$. – Jonas Meyer Jan 17 '11 at 6:15
@student: Yes, of course: Any neighborhood of any point $p$ of $\beta X$ contains points of $X$. But any point of $T$ is a point of $\beta X$, and any neighborhood $U$ of a point of $T$ (seen as a subset of $T$) is the restriction of a neighborhood $V$ of that point (seen as a subset of $\beta X$). This gives us that $V\cap X\ne\emptyset$. But $X\subset T$, so $V\cap X=V\cap X\cap T=(V\cap T)\cap X=U\cap X$. – Andrés Caicedo Jan 17 '11 at 6:16

The answer to this question (also about Stone-Cech compactification) is similar.

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