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As we went over in the limit definition post, whenever x is within $\delta$ of p, f(x) is within $\epsilon$ of $f(p)$. Balls are generalizations of this. A ball of radius $\delta$ centered at $p$ is defined by
$$B(p, \delta) = \{x \in \mathbb{R}^n \mid ||x-p|| < \delta\}$$
So, this is a higher dimensional analog of a delta-range or an epsilon-range.

Open Sets

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What is an open set? Just think of it as a region with no boundary. That means for every point, there exists a tiny region around that point that's still in the region.
For example, if a point is 1 unit from the "boundary", then all points 0.5 units from the point are inside the region. That gives us our mathematically accurate definition.
$$\text{A subset U}\subseteq \mathbb{R}^n \text{ is open means } \forall p \in U, \exists \delta \gt 0 s.t. B(p, \delta) \subseteq U$$

Examples

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Let $S = \{x \in \mathbb{R}^2 \mid x_1 > 0\}$. This is the right portion of a plane.
We want to prove for every point, there exists a ball that is still in the set.
$$\text{Let }a \in S$$
Now, we want to show that a ball exists for this point, so we pick a radius called $\delta$.
$$\text{Let }\delta = \dfrac{a_1}{2}$$
$$\text{Show: }B(a, \dfrac{a_1}{2}) \subseteq S$$
$$\sqrt{(x_1 - a_1)^2 + (x_2 - a_2)^2} \lt \dfrac{a_1}{2}$$
$$(x_1 - a_1)^2 + (x_2 - a_2)^2 \lt \dfrac{a_1^2}{4}$$
$$(x_1 - a_1)^2 \lt \dfrac{a_1^2}{4}$$
$$x_1^2 - 2a_1x + a_1^2 \lt \dfrac{a_1^2}{4}$$
$$x_1^2 - 2a_1x \lt \dfrac{-3a_1^2}{4}$$
$$x_1^2 - 2a_1x \lt 0$$
$$x_1^2 \lt 2a_1x$$
This is only true when $x_1 > 0$. Therefore, for every point in $S$, there exists a ball (radius = $\frac{a}{2}$) such that the ball is inside S. Therefore, S is open.

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We want to prove this for an arbitrary point, so we say $$\text{Let } y \in B(x_0, r)$$
Now, we need to find a delta that will work for this point. Let's say it's the distance to the edge, meaning $r - ||y - x_0||$
$$\text{Let } \delta = r - ||y-x_0||$$
Now, we want to show that the ball with that radius is contained inside the original ball, so we choose an arbitrary point in that ball and show it's in the original ball.
$$\text{Let } x \in B(y, \delta)$$
$$||x - y|| < r - ||y - x_0|| \text{ By definition of ball}$$
$$||x - y|| + ||y - x_0|| < r \to ||x - x_0|| < r $$
$$ x \in B(x_0, r)$$
Therefore, $$\forall y \in B(x_0, r), \exists \delta s.t. B(y, \delta) \in B(x_0, r)$$ This means it's open.