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Let $f$ be a Borel measurable function from a separable metric space $X$ onto a metric space $S$ with metric $e$. Show that $(S, e)$ is separable. Hints: As in Problem 8, $X$ has at most $c$ Borel sets. If $S$ is not separable, then show that for some $\varepsilon>0$ there is an uncountable subset $T$ of $S$ with $d(y, z)>\varepsilon$ for all $y = z \in T$ . Use Problem 9 to get a measurable function $g$ from $X$ onto $T$ . All $g^{−1}(A)$, $A \subset T$ , are Borel sets in $X$.

Assuming I have done all that, I just cannot find the contradiction to finish the proof. As I understand we have

  1. An injection $I_1:\mathcal B(X)\to \mathbb R$.

  2. An injection $I_2:\mathcal P (T)\to \mathcal B(X)$.

But how to finish? Do we need the continuum hypothesis?

EDIT: The continuum hypothesis seems sufficient, but is there a way to avoid it?

Alphie
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  • The contraction will be in the fact that the cardinality of $\mathcal{P}(T)>\mathcal{c}$. The sets $g^{-1}(A)$, $A\subset T$ are all measurable subsets of $X$. – Mittens May 01 '21 at 19:46
  • @OliverDiaz But how to show that the cardinality of $\mathcal{P}(T)$ is strictly greater than $\mathcal{c}$? – Alphie May 01 '21 at 19:47
  • $T$ is uncountable. Cantor-Berstein's theorem states that the cardinality of the power set of a set $B$ is striclty larger than the cardinality of the set $B$. – Mittens May 01 '21 at 19:53
  • @OliverDiaz This tells me that $\mathcal{P}(T)$ has cardinality strictly greater than $T$, but is the cardinality of $T$ greater than or equal to $c$? – Alphie May 01 '21 at 19:58
  • If T is uncountable, then it’s cardinal it’s is at least that of the continuum ( I don’t mind believing in the continuum hypothesis as being true, same with the axiom of choice) – Mittens May 01 '21 at 20:11
  • @OliverDiaz I see but are there additional assumptions that could avoid the use of the continuum hypothesis for this result? – Alphie May 01 '21 at 20:18
  • Like Said, If $T$ is constructed (in some other exercise) where once can easily see that its cardinality is that of $\mathbb{R}$ (or larger) then done. Otherwise, there is not other thing I can see to play with. – Mittens May 01 '21 at 20:29
  • I am looking at my old edition of Dudley's book (the one where strangely two pages from the subject index never made it for printing: I thought I was my copy was defective, but I learnt that was an error in all printings) Any way, in exercises 8 through 10, Dick was not timid about the HC, but it seems that in Exercise 8, the set $T$ I explicitly shown to have cardinality at least $c$. – Mittens May 01 '21 at 20:33
  • @OliverDiaz Could you show the statement of exercise 8? – Alphie May 01 '21 at 20:41
  • In exercise 8, Dudley is dealing directly with $\mathbb{R}$, so there it works.Since $X$ is a general separable metric space, then its cardinality is at most $c$. The same argument used in 8 would work ( I don't want to go into the guts of set theory again, try the follow the argument given as hint). The point is that the cardinality of Borel sets in $X$ is also $c$. For the last argument, the way I se it is basically continuum hypothesis: the smaller cardinal number that is uncouble is $c$, so $T$ being uncountable has to have cardinality larger or equal to $c$. – Mittens May 01 '21 at 21:01
  • Look here and there – Mittens May 01 '21 at 21:02
  • I just realize that you may do without HC (I think): $\kappa:=\operatorname{car}(T)>\aleph_0$ and so $2^\kappa>2^{\aleph_0}=c$. – Mittens May 01 '21 at 21:28
  • @OliverDiaz Yes I agree that $\operatorname{car}(T)>\aleph_0$, but we could have $2^\kappa=2^{\aleph_0}=c$ no? See here:https://math.stackexchange.com/a/376540/522332 – Alphie May 01 '21 at 21:47
  • Well, then HC, but you seem not to like the idea. – Mittens May 01 '21 at 21:56
  • Indeed its just that am a bit disappointed that the argument comes down to this assumption. If its the only way then so be it. But I believe the result might hold without this assumption if $X$ is Polish. – Alphie May 02 '21 at 02:50
  • Nothing wrong with that in my opinion. At the end of the they you are still using the AC, so using these two assumptions will not do harm. Again, if the case of cards fall with AC+CH it will also fall without them. Most remarkable results in Analysis require both and the say has not fallen yet. – Mittens May 02 '21 at 03:22

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