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There is a nice property of normal spaces, namely, closed disjoint subsets can be separated by continuous functions into $\mathbb R$.

Then you ask yourself, what about Hausdorff spaces?, are all Hausdorff spaces completely Hausdorff?

Well, if we choose our separating space $X$, that is, the space such that for any Hausdorff $Y$ and any distinct $a,b\in Y$ there is some continuous $f:Y\rightarrow X$ with $f(a)\neq f(b)$, to be completely regular and Hausdorff, this cannot happen by an example of E. Hewitt, Hewitt’s Condensed Corkscrew, call it $\mathbb H$; as any continuous function from $\mathbb H$ into $\mathbb R$ is constant, and any Hausdorff completely regular space $X$ can be embedded into a product of closed intervals $\prod_{i\in I}[0,1]$, and thus any continuous $f:\mathbb H\rightarrow X\subseteq \prod_{i\in I}[0,1]$ is constant, as each $\pi_i\circ f$ is constant, so our candidate to be the separating space $X$ cannot even be $[0,1]$.

My question is, does there exist a Hausdorff space $X$ such that for all Hausdorff spaces $Y$, and any distinct $a,b\in Y$ there exists a continuous $f:Y\to X$ such that $f(a)\neq f(b)$?

Thanks.

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    $\begingroup$ Completely normal is already used to denote a space where each subspace is normal. This is just because, as you mentioned, a space where two disjoint closed sets can be separated by a functions already falls under the term normal space. $\endgroup$ Commented Aug 11, 2013 at 19:57
  • $\begingroup$ Every completely regular Hausdorff space can be embedded into a product of closed intervals, so the space $X$ can't be completely regular. I assume that's what you meant? $\endgroup$
    – dfeuer
    Commented Aug 12, 2013 at 21:36
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    $\begingroup$ The accepted answer to this old question of mine is probably relevant: math.stackexchange.com/questions/72442/… $\endgroup$
    – dfeuer
    Commented Aug 12, 2013 at 21:40

1 Answer 1

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There does not exist such $X$.

Given any Hausdorff space $X$, there exists a Hausdorff $Y$ and distinct $a,b\in Y$ such that each continuous $f\colon Y\to X$ satisfies $f(a)=f(b)$.

This follows from the following more precise statement, which I will prove.

Let $X$ be Hausdorff with cardinality $\kappa=\lvert X\rvert$, and $\kappa^\prime$ be a cardinal such that there is no finite-to-one mapping $\kappa^\prime\to\kappa$. Then, there exists a Hausdorff space $Y$ with distinct points $a,b\in Y$ and subset $Z\subseteq Y$ satisfying

  1. Every closed neighborhood of $a$ or $b$ contains all but finitely many points of $Z$.
  2. $Z$ has cardinality $\kappa^\prime$.

Furthermore, for any such $Y$, every continuous $f\colon Y\to X$ satisfies $f(a)=f(b)$.

The statement that there is no finite-to-one mapping $\kappa^\prime\to\kappa$ just means that there is no function $g\colon\kappa^\prime\to\kappa$ with finite inverse images $g^{-1}(k)$ for all $k\in\kappa$. It can be seen that $\kappa^\prime=\aleph_0^\kappa$ works.

I'll first show that if $Y$ is chosen as above then all continuous $f\colon Y\to X$ satisfy $f(a)=f(b)$. Then, I'll show how to construct $Y$.

So, suppose that $f\colon Y\to X$ is continuous with $f(a)\not=f(b)$. As $X$ is Hausdorff, there are disjoint open neighbourhoods $U,V$ of $f(a),f(b)$. Also, as $Z$ has cardinality $\kappa^\prime$, there exists a point $x\in X$ with $f^{-1}(x)$ containing an infinite subset of $Z$. As the closure of $f^{-1}(V)$ contains all but finitely many points of $Z$, there exists a point $c$ in the intersection of $Z$, $f^{-1}(x)$ and the closure of $f^{-1}(V)$. In particular, $c$ cannot be contained in $f^{-1}(U)$, otherwise $f^{-1}(U)\cap f^{-1}(V)=f^{-1}(U\cap V)$ would be nonempty. This means, in particular, that $f(a)\not= f(c)=x$. Then, as $X$ is Hausdorff, there exist disjoint open neighbourhoods $U^\prime,W$ of $f(a)$ and $x$. Then, $f^{-1}(U^\prime)$ is an open set containing $a$ whose closure is contained in $Y\setminus f^{-1}(W)\subseteq Y\setminus f^{-1}(x)$, contradicting the fact (from 1) that it contains all but finitely many points of $Z$. QED

Now, I'll construct the Hausdorff space $Y$. Let the set of points in $Y$ be the collection of the following distinct points.

  • $a$ and $b$.
  • $a_{k,n}$ and $b_{k,n}$ for $k\in\kappa^\prime$ and $n\in\mathbb{N}$.
  • $c_k$ for $k\in\kappa^\prime$.

Also, let $Z=\lbrace c_k\colon k\in\kappa^\prime\rbrace$, which has cardinality $\kappa^\prime$. Let $\mathcal U$ be the collection of subsets of $Y$ of the following form.

  1. $\lbrace a\rbrace\cup\lbrace a_{k,n}\colon k\in\lambda,n\in\mathbb{N}\rbrace$ for any cofinite subset $\lambda$ of $\kappa^\prime$.
  2. $\lbrace b\rbrace\cup\lbrace b_{k,n}\colon k\in\lambda,n\in\mathbb{N}\rbrace$ for any cofinite subset $\lambda$ of $\kappa^\prime$.
  3. $\lbrace a_{k,n}\rbrace$ for any $k\in\kappa^\prime$ and $n\in\mathbb{N}$.
  4. $\lbrace b_{k,n}\rbrace$ for any $k\in\kappa^\prime$ and $n\in\mathbb{N}$.
  5. $\lbrace c_k\rbrace\cup\lbrace a_{k,n}\colon n\ge N\rbrace\cup\lbrace b_{k,n}\colon n\ge N\rbrace$ for any $k\in\kappa^\prime$ and $N\in\mathbb{N}$.

It can be seen that the intersection of any two elements of $\mathcal{U}$ is a union of elements of $\mathcal{U}$, so it defines a topology for $Y$, which it can be checked is Hausdorff. Finally if $U$ is a neighbourhood of $a$ or $b$, then it contains a set of the form (1) or (2) for some cofinite subset $\lambda$ of $\kappa^\prime$ and, hence, its closure contains the cofinite subset $\lbrace c_k\colon k\in\lambda\rbrace$ of $Z$.

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