# Show a set is a closed set larger than open set

let $(\mathbb{R}^n, \tau)$ where $\tau$ is the std topology. Show that the set $U = D_{\delta}(x) = \{y\in \mathbb{R}^n: ||y-x||\leq \delta \}$ is a closed set.

so under these conditions it means I have to show that $\mathbb{R}^n \setminus U$ is open

So the first thing to consider is what does an open set look like:

$\mathbb{R}^n \setminus U$ = $D_{\delta}(x) = \{y\in \mathbb{R}^n: ||y-x||> \delta \}$

proof:
let $\varepsilon = ||x-y|| - \delta$, let $z\in D_{\epsilon}(x)$,

want to show: $D_{\delta}(x) \subset D_{\epsilon}(x)$

$||z-x|| = ||z - y + y - x|| \leq ||z-y|| + ||y-x|| < \epsilon + \epsilon + \delta$

Problems: Now I know that I have to manipulate my inequality to some how get $\delta < ||z-x||$ but I can't figure out how to get it to work out. Suggestions?

You are right in that you have to show the complement is open. To do this, you select an arbitrary point in the complement and show that there is an $\varepsilon$-neighborhood of the point so that the neighborhood is contained entirely within the complement. Here, the choice of $\varepsilon$ can be dependent on your point, unlike how you have chosen your $\varepsilon$ (I can tell you were getting at that point, however!).

Now we can begin the proof. The intuition behind my choice of $\varepsilon$ is the fact that the point is not in the neighborhood, i.e. there is some 'wiggle-room' around the point where we can move and still be in the complement. This 'wiggle-room' will be our $\varepsilon$. More precisely, let $z \in \mathbb{R}^n \setminus U$. Since $z \notin U$, $||z - x|| > \delta$. Set $\varepsilon = ||z - x|| - \delta > 0$. Now we show $N_\varepsilon(z) \subset (\mathbb{R}^n \setminus U)$. Let $w \in N_\varepsilon(z)$. We need to show $w \in (\mathbb{R}^n \setminus U)$, i.e. $||x - w|| > \delta$. To this end, we will use the triangle inequality.

$$||x - z|| = ||x - w + w - z|| \leq ||x - w|| + ||w - z||$$ $$||x - w|| \geq ||x - z|| - ||w - z|| > ||x - z|| - \varepsilon = ||x - z|| - ||z - x|| - \delta = \delta,$$

where we heavily use the fact that $||w - z|| < \varepsilon$.

I hope the motivation for the choice of $\varepsilon$ was clear -- traditionally, this is done by drawing a picture for basic facts of point-set topology.

• yea I had the picture drawn, just couldn't phrase it right. I also negelceted to use the other form of the triangle inequality – dc3rd Jun 18 '15 at 1:04

You're getting a bit mixed up on the notation, I think. You want $z\in D_{\varepsilon}(y)$, then by the triangle inequality, $$\|x-y\| \leqslant \|y-z\| + \|z-x\|.$$Since $\|x-y\|=\delta+\varepsilon$ and $\|y-z\|\leqslant\varepsilon$, we have $$\|z-x\| \geqslant \|x-y\|-\|y-z\| \geqslant\delta+\varepsilon - \varepsilon = \delta,$$ so that $z\in\mathbb R^n\setminus U$.

• I also didn't use the othe form of the triangle inequality. – dc3rd Jun 18 '15 at 1:03