Arithmetical hierarchy and complexity course note? In my note, our professor talk about Arithmetical hierarchy. at the end he wrote all of these is True. My main problem is how these are True? ($N$ means Natural Numbers, and $W_e$   is the set of inputs for which the program encoded by $e$  halts)

{$ x \in \mathbb{N}| W_x$ is recursive } $ \in \Sigma_3$
{$ x \in \mathbb{N} | W_x$ =$N$ } $ \in \Pi_2$
{$ x \in \mathbb{N} | \bar{W}_x$ is finite } $ \in \Sigma_3$

Thanks to any idea or hint. 
Edit after 1 answer is received. infact when we want to solve this question via the proposed answer we get stuck when we ran into:

{$ x \in \mathbb{N} | \phi_x$ is total and bounded} $\in \Pi_2$

 A: $\varphi_x$ is total and bounded if and only if it satisfies the following two conditions:
(1) $\forall a\exists b (\varphi_x(a)[b]\downarrow)$ (totality; here . . . $[b]$ means "in $b$ steps"); and
(2) $\exists c\forall a(\varphi_x(a)<c)$ (boundedness).
Now (1) is $\Pi^0_2$, but (2) is $\Sigma^0_2$; so "total and bounded" = "(1) and (2)" is not obviously $\Pi^0_2$.
In fact, the statement is wrong - the right claim should be "total and unbounded is in $\Pi^0_2$" (and, in fact, is $\Pi^0_2$-complete).

Let's show that, in fact, $Y$="total and bounded" is not in $\Pi^0_2$. How would we do this? 
Well, a set $A\subseteq\mathbb{N}$ is $\Pi^0_2$ if and only if there is some computable total function $f(x, s)$ of two variables such that $$n\in A\iff\lim \sup_{s\rightarrow\infty} f(n, s)=\infty.$$ So suppose $f(x, s)$ is a computable total function; we'll find some $y$ such that $$y\in Y\iff\lim \sup_{s\rightarrow\infty} f(n, s)\not=\infty.$$  This will show that $Y$ is not $\Pi^0_2$.
This turns out to actually be rather easy: let $y$ be such that $\varphi_y(n)=f(y, n)$. (Exercise: how do we know that such a $y$ exists? This requires a theorem . . .) Then it's clear that $\varphi_y$ is total, so $y\in Y$ iff $\varphi_y$ is bounded.
BUT! $\varphi_y$ is bounded iff the function $s\mapsto f(y, s)$ is bounded, so we have $y\in Y\iff\lim\sup_{s\rightarrow\infty}f(y, s)\not=\infty$, so we are done.
