I want to find $$\bigcup_{n=1}^{\infty}\bigcap_{t \in [n, +\infty)}[t,t^2]$$ and $$\bigcap_{n=1}^{\infty}\bigcup_{t \in [n, +\infty)}[t,t^2]$$ My attempt. For the first one, we notice $$\bigcap_{t \in [n, +\infty)}[t,t^2] = [n,n^2] \cap [a_1,a_1^2] \cap [a_2,a_2^2] \cap \dots$$ At the same time $$[n,n^2] \cap [n^2+1,(n^2+1)^2] = \emptyset$$ This means $$\bigcap_{t \in [n, +\infty)}[t,t^2] = \emptyset$$ and $$\bigcup_{n=1}^{\infty}\bigcap_{t \in [n, +\infty)}[t,t^2]=\emptyset$$ For the second one $$\bigcup_{t \in [n, +\infty)}[t,t^2] = [n,n^2] \cup [a_1,a_1^2] \cup [a_2,a_2^2] \cup \dots = [n, +\infty)$$ Now, we have $$\bigcap_{n=1}^{\infty} [n, +\infty)= [1, +\infty) \cap [2, +\infty) \cap [3, +\infty) \cap \dots$$

Does this mean $\bigcap_{n=1}^{\infty}\bigcup_{t \in [n, +\infty)}[t,t^2] = \emptyset$?

  • $\begingroup$ To me, everything is ok, but the last statement. It seems to me that if we consider $R$ with $\infty$, than it has in some sense a unique element $\infty$(even though none of these sets contains $\infty$). $\endgroup$ – kolobokish Dec 8 '16 at 11:31
  • $\begingroup$ And one more thing. To prove your first statement as it is said in the answer beneath, you should show that for every $n$ , $n+1 \in (n,n^{2})$. $\endgroup$ – kolobokish Dec 8 '16 at 11:34

For the first one, your result is correct, but your method is flawed.

You claim

$$\bigcap_{t \in [n, +\infty)}[t,t^2] = [n,n^2] \cap [a_1,a_1^2] \cap [a_2,a_2^2] \cap \dots$$

But this is highly inacurate for two reasons:

  1. You did not define what $a_i$ is.
  2. The intersection is indexed with an uncountable set, so it's not accurate to write it down the way you did.

What you can claim is that if $x\in \bigcap_{t \in [n, +\infty)}[t,t^2]$, then, for every $t\in [n,\infty)$, $x$ must be an element of $[t,t^2]$.

This should be enough to show that the intersection is $\emptyset$, and from there, that the entire set is the empty set.

For the second, you made the same mistake with $a_i$ as in the first.

Also, you are rushing and are making claims without proof.

Your intuition is correct that $$\bigcup_{t\in [n, \infty)}[t,t^2] = [n,\infty),$$ but this is math. It isn't enough to just think that some equation must hold, you have to prove it does.

To prove the statement, you have to prove that the two sets are equal, and the easiest way to show this is to show that each is a subset of the other. In other words, show that

  • If $x\in \bigcup_{t\in [n, \infty)}[t,t^2]$, then $x\in [n,\infty)$.
  • If $x\in [n,\infty)$, then $x\in \bigcup_{t\in [n, \infty)}[t,t^2]$.

Again, note that $x\in \bigcup_{t\in [n, \infty)}[t,t^2]$ if and only if, for some $t\geq \infty$, $x$ is an element of $[t,t^2]$


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