Ytukyg

For $ n,m in Bbb N $, $A_{n,m}=left{ x in Bbb R: n^2 le x<m^2+(n+1)^2 right}$, find
$$bigcap_{m=0}^{infty} bigcup_{n=0}^{infty} A_{n,m}$$
$$bigcup_{m=0}^{infty} bigcap_{n=0}^{infty} A_{n,m}.$$

I suppose that $bigcap_{m=0}^{infty} bigcup_{n=0}^{infty} A_{n,m}=R^{+}$
and $bigcup_{m=0}^{infty} bigcap_{n=0}^{infty} A_{n,m} = emptyset $.

First result: For any $n, m in mathbb{N}$,

$$[n^2, (n+1)^2[subseteq A_{n,m}$$

$$implies bigcup_{n=0}^{infty}[n^2, (n+1)^2[ subseteqbigcup_{n=0}^{infty} A_{n,m}$$
$$implies R^{+} subseteqbigcup_{n=0}^{infty} A_{n,m}$$
$$implies bigcup_{n=0}^{infty} A_{n,m} = R^{+}$$ for all $m$. Then you intersect $R^{+}$ with itself for all $m$ and you get your first result.

Second result: We take a given $m in mathbb{N}$. We want to find $n_1, n_2 in mathbb{N}$ such that $A_{{n_1},m} bigcap A_{{n_2},m} = emptyset$. All you have to do is pick $n_1 = 0$ and $n_2 = m+1$ : $$A_{{0},m} bigcap A_{{m+1},m} = [0, m^2 + 1[ bigcap [(m+1)^2, m^2 + (m+2)^2[ = emptyset$$
The union of an infinity of empty sets is still an empty set so you get your second result.

We first show that any positive number belongs to $bigcap_{m=0}^{infty} bigcup_{n=0}^{infty} A_{n,m}$. Notice that any positive number lies always between two consecutive perfect squares i.e. $$forall x>0qquadexists Nqquad N^2le x<(N+1)^2$$therefore $$N^2le x<(N+1)^2le m^2+(N+1)^2$$which means that for all $min Bbb N^*$ we have $xin A_{n,m}$ for some $n$ therefore $xin B_m=bigcup_{n=0}^{infty} A_{n,m}$ and this means that $xin bigcap_{m=0}^{infty} B_m=bigcap_{m=0}^{infty} bigcup_{n=0}^{infty} A_{n,m}$. Also $0in bigcap_{m=0}^{infty} bigcup_{n=0}^{infty} A_{n,m}$ from which we obtain$$bigcap_{m=0}^{infty} bigcup_{n=0}^{infty} A_{n,m}=Bbb R^{ge 0}$$

Second, we show that no non-negative real belongs to $bigcap_{n=0}^{infty} A_{n,m}$ for all $m$. This can be simply proved since any real non negative number is finite and for large enough $n$ (i.e. $n^2>x$) we have $xnotin A_n$ which means that $bigcap_{n=0}^{infty} A_{n,m}=emptyset$. Since the union of empty sets is still an empty set we finally obtain$$bigcup_{m=0}^{infty} bigcap_{n=0}^{infty} A_{n,m} = emptyset $$and the proof is complete.

Let $xin mathbb R^{+}$ and $ mgeq 0$. Take $n=[sqrt x]$ and verify that $n^{2}leq x leq m^{2}+(n+1)^{2}$. This proves the first relation. Suppose the $y$ belongs to the second set. Then there exists $m$ such that $n^{2}leq x leq m^{2}+(n+1)^{2}$ for all $n$. But then $n^{2} leq x$ and $xleq m^{2}+(1+1)^{2}$, so $n^{2} leq m^{2}+(1+1)^{2}$ for any $n geq 0$ which is absurd. Hence the second set is empty.

$$B = bigcap_{m=0}^{infty} bigcup_{n=0}^{infty} A_{n,m}$$

Since $A$ is never has a negative, you can reason that $B$ doesn't either. Proving the rest is equivalent to

$$(forall x in mathbb R^{ge 0} ~forall m in mathbb N ~ exists n in mathbb N)~
n^2 le x < (n+1)^2 + m$$

So to prove it you need to give me a function $F$, of the form $n = F(x,m)$, such that if I tell you $x$ and $m$ your function tells me which $n$ it is in. $F$ is pretty direct nothing tricky about it.

$$bigcup_{m=0}^{infty} bigcap_{n=0}^{infty} A_{n,m} = emptyset$$

$$(forall x in mathbb R) lnot (exists m in mathbb N ~forall n in mathbb N)~
n^2 le x < (n+1)^2 + m $$

equivalently

$$(forall x in mathbb R ~ forall m in mathbb N ~exists n in mathbb N)~
lnot (n^2 le x < (n+1)^2 + m)$$

So same thing again, your proof will be demonstrating a function $F$ such that $n = F(x, m)$, but in this case it is to give a contradiction of the inequality. If $x < 0$ it is easy,

$$n = begin{cases} x < 0 &quad text{Anything} \
x ge 0 &quad text{To be determined}
end{cases}$$

I suggest just pick an $n$ that is so big that $x$ can't be in range.