Why does a zero'th order quantifier not increase the complexity of a $\bf{\Sigma}^1_1$ or $\bf{\Pi{^1_1}}$ assertion?

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If $\mathcal{X}$ is a perfect product space, and $P\subset\mathcal{X}$ is a $\bf{\Sigma}^1_1$ or $\bf{\Pi}^1_1$ pointset, then why does adding a bounded existential quantifier keep it $\bf{\Sigma}^1_1$ (or $\bf{\Pi}^1_1$)?

To put some notation on it, let's just stick with $n=1$. So there's some closed set $F\subseteq\mathcal{X}\times\mathcal{N}$ such that for all x, $P(x)\iff\exists\alpha F(x,\alpha)$.

Why is the pointset $Q$ given by $\exists y P(x,y)$ also $\bf{\Sigma}^1_1$?