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I am no expert in descriptive set theory, but for some reason want to estimate the complexity of a certain formula.

The framework. Let be $F$ a projective set in the Polish space $X=\mathbb R^{\mathbb N}$, say it is in the family $\mathbf \Pi^1_n$. Then, for a fixed $x_0 \in X$ I consider the formula $\varphi$:

$$ x_0\cdot y \in F,$$

where multiplication is performed pointwise and $y$ is the free variable bounded by $X$.

Question. Is it possible to compute (or estimate) the complexity of $\varphi$ in terms of the projective complexity of $F$, that is, $n$?

This could well be obvious, however, it is not my area of expertise at all.

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I believe that the term arithmetic complexity is not used for projective sets such as the one above, because in the arithmetic hierarchy one is concerned with first order formulas on the language of arithmetic: individual variables range over the natural numbers and you have $+$, $\cdot$, $0$ and $1$ as available operations; notably quantifiers $\forall$ and $\exists$ can only be used with those individual variables. But in order to define a projective set, you need variables that range over (possibly uncountable) Polish spaces. This is better understood on the space $2^{\mathbb{N}} \cong \mathop{\mathrm{Pow}}(\mathbb{N})$, but of course there are Borel isomorphisms between this and any other uncountable Polish space (and this will not change the result, if we start at $\boldsymbol\Pi^1_1$, $\boldsymbol\Sigma^1_1$, or above).

A finer detail is that in the hypothesis you refer to a boldface $\boldsymbol\Pi^1_n$ set. Those, apart from the (second order) formula, need a parameter to be defined.

Now, let's consider the complexity of $$ Y := \{ y \in X : x_0 \cdot y \in F\}. $$

It is clear that if all components of $x_0$ are non null, then it will have exactly the same complexity as $F$ ($y\mapsto x_0\cdot y$ being a homeomorphism). As soon as some component is $0$, then $Y$ will be homeomorphic to a projection of $F$, and hence hence its complexity may grow to $\boldsymbol\Sigma^1_{n+1}$ (that is, one extra second order existential quantifier). Had we started with a $\boldsymbol\Sigma^1_n$ set, $Y$ would have been also $\boldsymbol\Sigma^1_n$.

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  • $\begingroup$ Thanks, that's very instructive. Just let me ask you if I understand this right. Suppose that $F$ is a projective filter (regarded as a subset of $2^{\mathbb N}$). For a variable $y\in \mathbb R^{\mathbb N}$ and $x_0\in \mathbb R$, I consider the formula $\lim_F y_n = x_0$. (This means, $\forall \varepsilon > 0 \exists A\in F \forall n\in A |y_n - x_0| < \varepsilon$. Is the complexity of the formula $\Pi^0_{n+2}$ if $F\in \mathbf \Pi^0_n$? That's because the set $A:=\{y\in \mathbb R^{\mathbb N}: \forall_m \{n\in \mathbb N: |y_n - x_0| < 1/m\}\in F$ will increase it by 2? $\endgroup$ – Tomasz Kania Apr 22 '20 at 8:07
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    $\begingroup$ Note that you are putting a $0$ superscript now; that would mean a Borel set. If we insist on $F\in \mathbf \Pi^1_n$, the set defined by that formula is a priori, indeed, $\boldsymbol\Pi^1_{n+2}$. But also note that you don't need all $\varepsilon\in\mathbb{R}$ to conclude, just those of the form $\frac{1}{n}$; and that's just a number quantifier. So using that, we are at $\boldsymbol\Sigma^1_{n+1}$ (because both pointclasses are closed under countable union and intersection). $\endgroup$ – Pedro Sánchez Terraf Apr 22 '20 at 16:37
  • $\begingroup$ (In case you really meant $0$ above, the definition would be just $\boldsymbol\Sigma^1_{1}$.) $\endgroup$ – Pedro Sánchez Terraf Apr 22 '20 at 16:38
  • $\begingroup$ Thanks, indeed, I meant $\mathbf \Pi^1_n$. So to make sure: I go from $\mathbf \Pi^1_n$ to $\mathbf \Sigma^1_{n+1}$ as I express the set whether the formula holds true by introducing one more universal quantifier? In such arguments, is it fine, or is it customary to write such sets as projections/complements of some products involving the original sets? This is more a question of what people working in the area consider standard. $\endgroup$ – Tomasz Kania Apr 22 '20 at 16:42
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    $\begingroup$ You go from $\boldsymbol\Pi^1_{n}$ to $\boldsymbol\Sigma^1_{n+1}$ if you add another existential quantifier (ranging over an uncountable Polish space) before the rest. (And from $\boldsymbol\Sigma^1_{n}$ to $\boldsymbol\Pi^1_{n+1}$ if you add a universal one.) In general, once you write out the definition it is clear enough what the complexity is. But you might recourse to a expression using set theoretical operations if it is not clear. Check Kechris' book for use cases. $\endgroup$ – Pedro Sánchez Terraf Apr 22 '20 at 16:50

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