3
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Wasn't Tychonoff $= T_{3\frac{1}{2}} + T_0$? Indeed. $\endgroup$ – Daniel Fischer Jun 24 '14 at 13:59
  • $\begingroup$ 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. $\endgroup$ – Mark Fantini Jun 24 '14 at 19:14
  • $\begingroup$ @Fantini I explained the last part as well. $\endgroup$ – Henno Brandsma Jun 24 '14 at 19:56
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I meant to ask if the image was dense but you explained it very well. Thank you! $\endgroup$ – Mark Fantini Jun 24 '14 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.