Proving two sets are equal Suppose we have a claim: $A = \bigcup_{k \in \mathbb N} [-k, \frac{1}{k}) = (-\infty,1)$ = B.
I aim to prove this by showing that the two sets are subsets of each other (i.e. $A \subset B$ and $B \subset A$).
As for my strategy of showing this, I am going to grab an arbitrary element from the set A and show that this also belongs to B and vice versa. 
Now, my problem is that although I can easily show this in picture, I lack the mathematical argument that describes this picture in my head. Any suggestions?
 A: To show that $A\subseteq B$, there is not much choice in the mathematical argument. Take an arbitrary element $a\in A$. What does this mean? This means that there exists a $k\in\mathbb{N}$ such that $a\in [-k,\frac{1}{k})$. But since $-\infty<-k$ and $\frac{1}{k}\leq1$, we have the inclusion $[-k,\frac{1}{k})\subseteq(-\infty,1)=B$. In particular, $a\in B$.
To show the opposite inclusion, take an arbitrary $b\in B$. Then $b$ is a real number in the interval $(-\infty,1)$. We have to show that there exists a $k\in\mathbb{N}$ such that $b\in[-k,\frac{1}{k})$. Now, we can distinguish three cases:


*

*If $b<0$, we know we can choose a $k\in\mathbb{N}$ such that $-k\leq b$ and such a $k$ does the job.

*If $b=0$, any $k\in\mathbb{N}$ does the job.

*If $b\in(0,1)$, then $k=1$ does the job, since in that case we have $\frac{1}{k}=1>b$.

A: Let $x \in B$. If $0 \leq x < 1$, then $x \in [-1,1)$ (plugging $k=1$), we get that $x \in A$. Now if $x<0$, $-x>0$. Pick an integer $k_{0} \in \mathbb{N}$ such that $-x<k_{0}$,hence $x \in [-k_{0},0) \subset[-k_{0},\frac{1}{k_{0}} )$. Hence $B \subseteq A$. Now let $y \in A$, then we can find $k_{1} \in \mathbb{N}$ such that $y \in [-k_{1},\frac{1}{k_{1}})$. hence $- \infty <k_{1}<y <\frac{1}{k_{1}}<1$ (although the first inequality is not proper to write, but its just to make it concrete), the last inequality is not hard to see!
A: Note first that for any $k\in\Bbb N,$ we have $$\frac 1k\le1,$$ and so $$\left[-k,\frac1k\right)\subseteq(-k,1)\subseteq(-\infty,1).$$ Since this holds for all $k\in\Bbb N,$ then $A\subseteq B.$
For the reverse inclusion, we have to resort to element-chasing. If $b\in B,$ we want to show that there is some $k\in\Bbb N$ such that $b\in\left[-k,\frac1k\right),$ meaning we need $-k\le b<\frac1k.$ Now, if $b\ge0,$ this is easy. If $b<0,$ we use the fact that $-b$ isn't an upper bound of $\Bbb N$ to deduce that there is some $k\in\Bbb N$ such that $-b\le k,$ and from there it's straightforward.
