1
$\begingroup$

Let $\mathbb P$ denote the space of irrational numbers. In an answer to this question, Taras Banakh showed that the perfect images of $\mathbb P$ are precisely the Polish spaces with no compact neighborhoods. Here, perfect means a continuous, closed, surjective mapping with compact point pre-images.

Increasing the dimension slightly, we go from $\mathbb P$ to complete Erdős space $$\mathfrak E_{\mathrm{c}}=\{x\in \ell^2:x_n\in \mathbb P\text{ for all }n<\omega\}.$$ Here, $\ell^2$ is the Hilbert space of square-summable sequences of real numbers.

Question 1. Is every perfect image of $\mathfrak E_{\mathrm{c}}$ homeomorphic to $\mathfrak E_{\mathrm{c}}$?

Question 2. Is $\mathbb P$ a perfect image of $\mathfrak E_{\mathrm{c}}$?

$\endgroup$
  • $\begingroup$ What about the original Erdos space? (Does everybody but I know?) $\endgroup$ – Wlod AA Apr 19 at 1:28
  • 2
    $\begingroup$ @WlodAA That one is usually more difficult to work with because it does not have as many representations, is not Polish, etc. But that may be a good follow-up question. $\endgroup$ – D.S. Lipham Apr 19 at 1:37
1
$\begingroup$

It seems that the answer to Question 1 is "no".

According to this paper, the Julia set of $f(z)=\pi\sinh(z)$ is equal to the entire complex plane $\mathbb C$, and is the perfect image of a "Cantor bouquet". The endpoint set of any Cantor bouquet is homeomorphic to $\mathfrak E_{\mathrm{c}}$. But according to the image below (from the same paper), these endpoints are mapped to a dendritic connected set (see the dark lines including the imaginary axis). enter image description here

| cite | improve this answer | |
$\endgroup$

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.