When defining Stone–Čech compactification we take a Tychonoff space $X$, the space $C_b(X)$ of bounded continuous real functions on $X$, define $I_f$ as closed limited intervals containing $f(X)$ for each $f \in C_b(X)$ and the mapping $$\varepsilon_X : \quad x \in X \mapsto (f(x))_{f \in C_b(X)} \prod_{f \in C_b(X)} I_f.$$ My notes assert that this mapping is an embedding by completely regular spaces properties. Defining $\beta X$ as $\beta X = \overline{\varepsilon_X (X)}$ we obtain the Stone–Čech compactificationwith the pair $(\beta X, \varepsilon_X)$.

To have a compactification we need that $X$ is a Hausdorff space, $\beta X$ be a compact Hausdorff space and $\varepsilon_X$ be a homeomorphism between $X$ and a dense subspace of $\beta X$. My questions are:

  • $X$ is merely Tychonoff and not Hausdorff, how does this work then?
  • How can we assert that $\varepsilon_X$ is homeomorphic to a dense subspace of $\prod I_f$?

We know that it is homeomorphic to its image, but is the image dense? How and why?

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    $\begingroup$ Wasn't Tychonoff $= T_{3\frac{1}{2}} + T_0$? Indeed. $\endgroup$ Jun 24, 2014 at 13:59
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    $\begingroup$ The material I'm using defined Tychonoff as $T_1$ + completely regular, where $T_1$ means that for every two distinct points you can find one open set for each that doesn't contain the other. Since that didn't have Hausdorff nor I saw how it could imply it left me wondering. $\endgroup$ Jun 24, 2014 at 19:14
  • $\begingroup$ @Fantini I explained the last part as well. $\endgroup$ Jun 24, 2014 at 19:56

1 Answer 1


In this theorem Tychonoff should imply Hausdorffness (so the function separation property plus $T_0$ (or equivalently $T_1$ or $T_2$)). This is needed as we want to embed $X$ into a compact Hausdorff space so it should certainly be Hausdorff to start with, as subspaces of Hausdorff spaces are Hausdorff.

It does not assert it embeds as a dense subspace of $\prod_f I_f$. As $\epsilon_X$ is an embedding into $\prod_f I_f$, we know that $X$ is homeomorphic to $\epsilon_X[X]$. And the Čech-Stone compactification $\beta X$ is defined as $\overline{\epsilon_X[X]}$ (closure taken in $\prod_f I_f$) so by definition $X$ (in the form of the homeomorphic $\epsilon_X[X]$) is by definition dense in it (as its closure is exactly that space). And this closure is of course compact Hausdorff as a closed subset of a compact Hausdorff space $\prod_f I_f$. $\epsilon_X$ is a function, so it's not dense in anything...

To address your concern in the comments: if $X$ is $T_1$, this means points (really, singletons) are closed. So if $x \neq y$ are points in $X$ we apply complete regularity to $x$ and the closed set $\{y\}$ to get a continuous function $f: X \rightarrow [0,1]$ such that $f(x) = 0$ and $f(y) = 1$. Then $f^{-1}[[0,\frac{1}{2})]$ and $f^{-1}[(\frac{1}{2}, 1]]$ are open disjoint neighbourhoods of $x$ and $y$ respectively (inverse images of disjoint open sets in $[0,1]$ under a continuous function..). This should have been a basic lemma in your text I think.

  • $\begingroup$ I meant to ask if the image was dense but you explained it very well. Thank you! $\endgroup$ Jun 24, 2014 at 19:15

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