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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?

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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.

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  • $\begingroup$ yea I had the picture drawn, just couldn't phrase it right. I also negelceted to use the other form of the triangle inequality $\endgroup$ – dc3rd Jun 18 '15 at 1:04
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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 $z\in\mathbb R^n\setminus U$.

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

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