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I want to prove the following:

For any set $X$ there exists a Hausdorff space $Y$ and a family $\{Y_x:x\in{X}\}$ of disjoint subsets of $Y$, each dense in $Y$.

I want to prove this assertion since by proving it, I can use it to prove that any topological space is the open continuous image of a Hausdorff space.

I have proven the assertion, but using mathematical logic; namely, using the compactness theorem and the fact that the assertion is true for finite $X$ in some ordered set, but I don't know how else to prove this without the use of the compactness theorem.

I would thank very much to see a prove of this, without using mathematical logic.

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up vote 4 down vote accepted

I believe you can actually exemplify such a space.

  1. Choose some arbitrary set $X$.
  2. Take the space $Y=X^\omega$ with each copy of $X$ with discrete topology, and $Y$ with the usual Tychonoff product topology. Then $Y$ is not only Hausdorff, but completely regular (as a product of completely regular spaces).
  3. For each $x\in X$ let $Y_x$ be the set of all sequences with all but finitely many terms equal to $x$.
  4. It is easy to see that $Y_x$ are disjoint and each is dense.

Better yet, I think you can choose $Y$ to be compact Hausdorff, though the construction is more combinatorially involved.

  1. Choose any infinite $\kappa$ such that $2^\kappa\geq\lvert X\rvert$ (for instance, $\kappa=\lvert X\rvert$).
  2. Put $Y=2^\kappa$ with product topology. $Y$ is compact Hausdorff by Tychonoff's theorem.
  3. Put on $Y$ an equivalence relation $\sim$ defined by $\sigma\sim \tau$ iff $\sigma$ and $\tau$ agree on all but finitely many axes.
  4. Notice that each class of $\sim$ is of cardinality $\kappa<2^\kappa$ (because there are only $\kappa$ finite subsets of $\kappa$), so there are $2^\kappa\geq \lvert X\rvert$ of them.
  5. Assign to each $x\in X$ a distinct equivalence class $Y_x$ of $\sim$.
  6. $Y_x$ are disjoint by the definition, and they are easily shown to be dense.

Edit: as suggested by Nate Eldredge, there is another, simpler way to obtain a compact Hausdorff example:

  1. Take for $\tilde X$ a compact Hausdorff space of size at least $\lvert X\rvert$ (for example, one-point or Čech–Stone compactification of $X$ with discrete topology, or $\lvert X\rvert+1$ as an ordinal with order topology).
  2. Take for $Y$ the product $\tilde X^\omega$; it is again compact Hausdorff.
  3. Assign to each $x\in X$ a distinct element $x'\in \tilde X$, and then put for $Y_x$ the set of all sequences in $Y$ with all but finitely many terms $x'$.
  4. Much like in the first example, $Y_x$ are disjoint and dense.

In hindsight, yet another way for a compact Hausdorff $Y$ would be to just take the Čech–Stone compactification of the $Y$ in the first example with the same $Y_x$ (but this one requires familiarity with the Čech–Stone compactification, so whether you consider it simpler or not is up to you).

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Nice example. I believe we can construct a compact Hausdorff example more easily. Just take $\tilde{X}$ to be any compact Hausdorff space of cardinality at least $|X|$; for instance, the Stone-Cech compactification of the discrete topology on $X$, or a large enough successor ordinal. Then take $Y = \tilde{X}^\omega$ and proceed as in your first construction. – Nate Eldredge Nov 24 '12 at 16:52
@NateEldredge: good point. Then again, now that I think of it, if you're willing to accept Čech–Stone, then you can just take the compactification of the first $Y$. :) – tomasz Nov 24 '12 at 17:43

If $|X|\le 2^\omega$ you can find such a space in the real line, the dense subsets being translates of $\Bbb Q$.

For more than that, let $\{Q_\xi:\xi<2^\omega\}$ be a partition of $\Bbb R$ into countable dense subsets, let $\kappa$ be a sufficiently large cardinal, let $X_\xi=\Bbb R$ for $\xi<\kappa$, and let $Y=\square_{\xi<\kappa}X_\xi$, the box product of the spaces $X_\xi$. For each $\varphi:\kappa\to 2^\omega$ let $Y_\varphi=\{y\in Y:y_\xi\in Q_{\varphi(\xi)}\text{ for all }\xi<\kappa\}$. The sets $Y_\varphi$ are pairwise disjoint and dense in $Y$, and there are $2^\kappa$ of them. Just choose $\kappa$ large enough so that $2^\kappa\ge|X|$ and ignore any extra dense sets.

More generally, start with any Hausdorff space $Z$ with disjoint dense subsets $D_0$ and $D_1$. For each $x\in X$ let $Z_x$ be a copy of $Z$, and let $Y=\square_{x\in X}Z_x$. For $x\in X$ let $$Y_x=\Big\{y\in Y:y_x\in D_1\text{ and }y_u\in D_0\text{ for all }u\in X\setminus\{x\}\Big\}\;;$$ then the sets $Y_x$ are pairwise disjoint, dense subsets of $Y$, which is Hausdorff.

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An answer using mathematical logic is as follows:

Let $L=\{<\}\cup{\{P_x:x\in{X}\}},$ where the $P_x$ are distinct unary predicates. For each $x\in{X}$ let $\varphi_x=\forall{y}\forall{z}(y<z\rightarrow\exists{w}(P_xw\wedge{y<w<z}))$ and for any $x,y\in{X}$ with $y\neq{x}$ let $\Delta_{x,y}=\forall{z}¬(P_xz\wedge{P_yz})$. Let $T=\{\varphi_x:x\in{X}\}\cup{\{\Delta_{x,y}:x,y\in{X}\wedge{x\neq{y}}\}},$ then $T$ is finitely satisfiable, since $(\mathbb{Q},<)$ can be partitioned in $n$ dense subsets for any $n<\omega$. Thus by the compactness theorem there exists a model $(M,<,P_x)_{x\in{X}}$ of $T$, so the $P_x^{\mathcal{M}}$ are dense disjoint subsets of $(M,<)$, but since order topologies are Hausdorff, $(M,<)$ is our required space.

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