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Let $C(X)$ be the set of all continuous, real-valued functions on a Tychonoff (completely regular) topological space $X$ and let $\beta X$ be the Stone-Cech compactification of $X$. Now let $f\in C(X)$ such that $f(X)=\{0, 1\}$ and $f^\beta$ be an extension of $f$ to $\beta X$, that is, $f^\beta\in C(\beta X)$. How can we show that $f^\beta(\beta X)=\{0, 1\}$?

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The Stone-Čech compactification has a very strong property: if $f\colon X\to Y$ is a continuous map with $Y$ compact, then $f$ can be (uniquely) extended to $\beta f\colon\beta X\to Y$. See Universal property and functoriality on Wikipedia.

Since $\{0,1\}$ is compact, you're done.

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$X$ is dense in $\beta(X)$. Hence $f^{\beta} (\beta(X)) = f^{\beta} (\overline {X}) $ which is contained in the closure of $\{0,1\}$ (by continuity) so it is contained in $\{0,1\}$. Reverse inclusion follows by the fact that $f^{\beta}$ extends $f$.

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