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I am trying to read Kechris book on Classical Descriptive Set Theory (self-studying only so please do bear with me). I need help in understanding the following part of the proof for (i) of Proposition 14.4.

From what I understand, a set $A$ is analytic if it is the projection of a closed set in $X \times \mathcal{N}$, where $X$ is Polish and $\mathcal{N}$ is the Baire space. However, $A$ by itself is not necessarily closed.

By a previous result, any Polish space $Y_n$ is homeomorphic with a closed subset of $\mathcal{N}$ - so we could suppose that $Y_n = \mathcal{N}$ for all $n$.

Now, in this proof, we have that $A_n = \{ f_n(x) : x \in \mathcal{N} \}$, so that $A_n$ is the projection of a closed set in $X \times \mathcal{N}$, i.e. the closed set $\{(f_n(x), x) : x \in \mathcal{N} \}$.

However, in this proof, we have that $Z$ is closed in $\Pi_nY_n$ (or in $\Pi_n \mathcal{N}$ if we take $Y_n = \mathcal{N}$ for all $n$) I don't follow why $Z$ is closed in this case. Any help ?

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Link L
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    In general it's true that if $Y$ is a topological space and $X$ is a Hausdorff topological space, and $f, g: Y \to X$ are continuous, then ${y : f(y) = g(y)}$ is closed in $Y$. In this case $Z$ can be written as an intersection of such sets (taking $f = f_n \pi_n$ and $g = f_m \pi_m$). I think such facts about topology are likely to be taken for granted in a descriptive set theory book - you should be able to recognise that the set $Z$ is defined by "closed conditions". – Izaak van Dongen Apr 12 '24 at 13:22
  • @IzaakvanDongen okay thanks, would you know of any reference where I could read that result ? – Link L Apr 12 '24 at 13:31
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    https://math.stackexchange.com/q/136922/473276 and https://math.stackexchange.com/q/2080863/473276 together would be the standard way to prove this fact (which is in fact equivalent to being Hausdorff). The really key part is that Hausdorff is equivalent to the diagonal being closed, which is often given as a lemma or exercise when you're first learning general topology. – Izaak van Dongen Apr 12 '24 at 13:49

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If $\upsilon^k=(y^k_n)_{n=1}^\infty\in Z$ converges to $\upsilon=(y_n)_{n=1}^\infty$, then for any $m,n\in\mathbb{N}$, since $f_n,f_m$ are continuous and $\lim_k y^k_n=y_n$ and $\lim_k y^k_m=y_m$, $$f_n(y_n)=\lim_k f_n(y_n^k)=\lim_k f_m(y_m^k)=f_m(y_m).$$

user469053
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  • ok, but to show that $Z$ is closed, I have to prove that given the condition $f_n(y_n)=f_m(y_m)$ for any $m,n \in \mathbb{N}$ (as per the construction of $Z$), we have to prove that the sequence $v^k \rightarrow v$ for some $v \in Z$ right ? – Link L Apr 12 '24 at 13:28
  • No. To show closedness, we want to show that $\overline{Z}\subset Z$. So we fix $\upsilon=(y_n)_{n=1}^\infty\in \overline{Z}$ and show that it's in $Z$ (by showing that it satisfies the condition $f_n(y_n)=f_m(y_m)$ for all $m,n\in\mathbb{N}$). But since we're working in a metric space, $\upsilon\in \overline{Z}$ iff there exists a sequence in $Z$ converging to $\upsilon$. So we pick such a convergent sequence, which must exist due to this characterization of $\overline{Z}$. – user469053 Apr 12 '24 at 14:52
  • In general, a subset of a metric space is closed iff it contains all the limits of its sequences. So to show closedness, it is sufficient to suppose that we have a sequence in the set converging to a point, and show that the point is also in the set. – user469053 Apr 12 '24 at 14:53