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For each positive integer, let $A_{k} = [\frac{1}{k},k+1)$ in the universe of all real number. I have to determine the sets and justify that it is correct for both $\bigcup_{k=1}^{\infty} A_{k}$ and $\bigcap_{k=1}^{\infty} A_{k}$.

All I really know for sure is that any number I put into the $[\frac{1}{k}, k+1)$ the $\frac{1}{k}$ approaches zero, while the $k+1$ approaches infinity. I'm not exactly sure how I'm supposed to determine the sets and justify that the answer is correct.

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    You question is hard to understand. What do you mean by "justify that it is correct"? Do you need to compute $\bigcup_{k=1}^\infty A_k$ and $\bigcap_{k=1}^\infty A_k$? – 2017-02-18
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    A good approach here is guess & verify. The fact that the sets $A_k$ are nested makes the problem a little simpler. – 2017-02-18
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    @JoshuaRuiter I'm not exactly sure what it means, to be honest. That's all that I was given. I'm assuming I have to write a proof in order to confirm the sets. – 2017-02-18

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Well, thinking about the intersection and union, the following conjectures should rise: $\bigcup_{k=1}^{\infty} A_{k}=(0,\infty)$ and $\bigcap_{k=1}^{\infty} A_{k}=[1,2)$.

First the union. Take a random $x\in(0,\infty)$. We should proof that $\exists j\in \mathbb{N}$, such that $x\in A_j$. Now, this is not hard to proof. If $x \geq1$, it suffices to take $j=\lceil x \rceil$ (actually, because of the $k+1$, $\lfloor x \rfloor$ is also sufficient). For $x<1$, we have to take $j=\lfloor x \rfloor$.

Now, can you proof the intersection yourself?

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    For the intersection, notice that if $k > n$ then $A_n \subset A_k$. – 2017-02-18