# Show that $\cap_{k\in \mathbb{N}}[-\frac{1}{k},k+1]$ is a closed set

I know that a subset $M$ of a metric space $(X,d)$ is open if it contains a ball about each of it points, and closed it its complement is open.

But how would I show that the set $\cap_{k\in \mathbb{N}}[-\frac{1}{k},k+1]$ in $(\mathbb{R},|.|)$ is closed and not open?

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Do you know in generality that the intersection of closed sets is closed? –  abatkai May 21 '12 at 20:40
The set has a much simpler description. Ask yourself: Which positive numbers are in the set? Which negative numbers are in the set? Is $0$ in the set? abatkai raises a good point, but in this case it is useful to have a more concrete description first. –  Jonas Meyer May 21 '12 at 20:42
Identify in a picture the intervals $[-1/k,k+1]$ for $k=1$, $2$, $3$, $4$. Find the intersection of these. –  André Nicolas May 21 '12 at 20:47
I was again too quick. @JonasMeyer is right. –  abatkai May 21 '12 at 20:47
Is the set between $[0,2]$? –  Steven May 21 '12 at 20:49

Let $C = \cap_{k\in \mathbb{N}}[-\frac{1}{k},k+1]$, and suppose $x \notin C$. Then $\exists k$, such that $x \notin [-\frac{1}{k},k+1]$. Choose $\epsilon = \frac{1}{2} \min \{ |x+\frac{1}{k}|, |x-(k+1)|\}$. Then $B(x,\epsilon) \cap C = \emptyset$, since $C \subset [-\frac{1}{k},k+1]$.
Thanks @copper.hat The only question is, how did you choose $\epsilon = \frac{1}{2} \min \{ |x+\frac{1}{k}|, |x-(k+1)|\}$? Is that a standard way to choose epsilon? –  Steven May 21 '12 at 20:55
Well, there is no standard way. In the above, either $x < -\frac{1}{k}$, or $x > k+1$. I just choose $\epsilon$ so it would be half the minimum distance from $x$ to the interval. Any other fraction would do, as indeed would many other schemes. –  copper.hat May 21 '12 at 20:59