I want to prove the following and have no idea how to proceed:

For a continuous function $f: X \mapsto Y$ where $X$ is Polish and $Y$ is Hausdorff the following are equivalent:

  1. $f[X]$ is uncountable

  2. There is a a perfect set $P \subset X$ such that $f|_p$ is one-to-one.

  3. There is a countable set $Q \subset X$ with no isolated points such that $f|_Q$ is one-to-one.

It 'smells' like the Cantor–Bendixson Theorem, but have no idea how to use it. Any help will be greatly appreciated.


This is as far as I can go in proving it. I will use the following theorems from Srivastava, A course on Borel sets (loosely cited):

2.6.1 Every dense-in-itself Polish space $X$ contains a copy of $2^\omega$

2.6.2 (Cantor – Bendixson theorem) Every separable metric space $X$ can be written as $X = P \cup Q$ where $Q$ is countable, $P$ is perfect, and $P,Q$ are disjoint.

2.6.7 Let $E$ be a closed equivalence relation on a Polish space $X$ with uncountably many equivalence classes. Then there is a copy of $2^\omega$ in X consisting of pairwise inequivalent elements.

2) $\Rightarrow$ 1)

There is perfect $P \subset X$ on which function is one to one. It contains homeomorph of $2^\omega$ and is separable, so $P$ is of cardinality $\mathfrak c$. Hence $|f[X]| \geq | f [P] | = |P| = \mathfrak c$

1) $\Rightarrow$ 2)

  1. By AC we choose one element from each nonempty layer $f^{-1}[\{y\}]$. We sum them up, acquiring uncountable $A \subset X$. Using C-B theorem we write it as $A = P \cup Q$ where $P$ is nonempty (because $A$ is uncountable and $Q$ is countable) set, perfect in $A$. Problem is that we are not sure if it is perfect in $X$. But $P$ is dense-in-itself, so if it was a $G_\delta$ set it would be a Polish space (Aleksandrov theorem). Hence it would contain a copy $D$ of $2^\omega$. Cantor set is perfect and $D$ is closed in $X$ by compactness hence it is perfect set in $X$, the one that we are looking for.

  2. Suppose that layers of $f$ induce closed equivalence relationship on $X$. Then there exists copy of Cantor set with one element in each equivalence class. By previous argument it is the perfect set we are looking for.

Unfortunately I have no idea how to assure the assumptions I am using. Any help/ideas?

  • $\begingroup$ This would have an interesting corollary: Every continuous function from a Polish space into $\omega_1$ (with the order topology) is bounded. Is this true? $\endgroup$ – Pedro Sánchez Terraf Jun 13 '15 at 19:33
  • $\begingroup$ I can't see how that corollary is implied by this theorem. What is true? This theorem was on my DST homework list, i suppose it is valid. $\endgroup$ – wroobell Jun 15 '15 at 20:37
  • $\begingroup$ Actually, what I said is not quite correct. But if CH is false it holds indeed, because $Y =\omega_1 $ is Hausdorff and any perfect subset of a Polish has cardinality of the continuum. Hence, the range can't be uncountable, and therefore it must be bounded. $\endgroup$ – Pedro Sánchez Terraf Jun 15 '15 at 20:44
  • $\begingroup$ Yes, now it makes sense. :) I think i have proved the equivalence of two first conditions, not that hard. I can post it tomorrow if You are interested. $\endgroup$ – wroobell Jun 15 '15 at 20:47
  • $\begingroup$ Please do so. I think that's the most difficult part. $\endgroup$ – Pedro Sánchez Terraf Jun 15 '15 at 23:54

Here's a sketch for (1) implies (2): Fix $A \subseteq X$ be uncountable such that $f \upharpoonright A$ is one-one. By deleting at most countably many points from $A$ we can assume that every point of $A$ is a condensation point of $A$ - Recall that $x$ is a condensation point of $A$ if every neighborhood of $x$ contains uncountably many points of $A$. Now define a tree $\langle B_{\sigma} : \sigma \in 2^{< \omega} \rangle$ such that the following hold: $B_{\phi} = X$, $B_{\sigma 0}, B_{\sigma 1}$ are disjoint closed balls contained in $B_{\sigma}$, $f[B_{\sigma 0}] \cap f[B_{\sigma_1}] = \phi$ and radius of $B_{\sigma} $ is at most $2^{-n}$. To guide this construction, always keep the centers of your balls in $A$. That $Y$ is Hausdorff helps you satisfy $f[B_{\sigma 0}] \cap f[B_{\sigma_1}] = \phi$. Finally your perfect set is simply all the points that lie in the intersection of a branch in $\langle B_{\sigma} : \sigma \in 2^{< \omega} \rangle$. It should be clear that $f$ is one-one on this perfect set.

Hope this helped :)

@Pedro's comment: You don't need $\neg CH$ to conclude that the range of any continuous function from a second countable space to $\omega_1$ is bounded. Just use the fact that every unbounded subset of $\omega_1$ contains uncountably many pairwise disjoint open sets.

  • $\begingroup$ I saw this by chance, and I'm glad of learning from your reply. Next time, try to answer (also) in the comment section, so that people gets noticed by your observations! $\endgroup$ – Pedro Sánchez Terraf Jun 29 '15 at 1:15
  • $\begingroup$ @PedroSánchezTerraf Unregistered users cannot comment except on their answers thanks to the rule-makers. $\endgroup$ – hot_queen Jul 3 '15 at 14:33
  • $\begingroup$ @hot_queen Oh, I see. Seems reasonable, that may induce people to register. Thanks. $\endgroup$ – Pedro Sánchez Terraf Jul 3 '15 at 18:25

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