How do I define the proper subset $\bigcup\limits_{i=1}^\infty[n^2,n^2+1]$ of a set $\bigcup\limits_{i=1}^\infty[n,n+1]$

Question

I have to determine if the set $T=\bigcup\limits_{i=1}^\infty[n^2,n^2+1]$ is bounded and find the supremum and infimum if they exist. Clearly, $T=\{[1,2], [4,5], [9,10]...\}$ it is clearly bounded below by 1 and has no upper bound. I want to prove this by showing that this is a proper subset of $A=\bigcup\limits_{i=1}^\infty[n,n+1]$ as this is clearly an easier set to work with. How do I prove that $A$ is a proper subset of $T$?

Answer 1

I am assuming you have meant $T= \bigcup_{n=1}^\infty[n^2,n^2+1]$, and you want to show that $A$ is a proper subset of $T$, not $B$ since there is not $B$ in your question. There is a confusion here, $T$ is not a family of intervals, it is a union of intervals. To show the inclusion,

Let $x \in T$, then there is $j\in \mathbb N$ such that $x\in [j^2,j^2 +1]$, since $j^2 = k \in \mathbb N$, then $x \in [k,k+1] \subset A$, thus $T \subset A$.

Now, take $3 \in A$, it is clear that $3 \in [2,3] \cap [3,4] \subset A$. But there is no interval in $T$ that contains $3$. Then $T$ is a proper subset of $A$.