Show that for each $c\in \mathbb{R}^n$ and $\delta >0$, the set: $V_g(c)= \{x\in \mathbb{R}^n:|x-c|< \delta\}$, is open.

Question

Show that for each $c\in \mathbb{R}^n$ and $\delta >0$, the set: $V_g(c)= \{x\in \mathbb{R}^n:|x-c|< \delta\}$, is open.

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My attempt:

$$\operatorname*{int}(V_g(c))= \{x\in \mathbb{R}^n: |x-c|< \delta\} $$ $$\operatorname*{ext}(V_g(c))= \{x\in \mathbb{R}^n: |x-c|> \delta\} $$ $$\operatorname*{bdy}(V_g(c))= \{x\in \mathbb{R}^n: |x-c|= \delta\} $$

I think to prove this I can show that there exists an open rectange in the exterior of $V_g(c)$ that doesn't touch the boundary and doesnt exist in $\operatorname*{int}(V_g(c))$. I am now sure how to go about this.

Answer 1

The first statement of your three is just a restatement of the fact that $V_g(c)$ is open, so it's begging the question (Asuming as true that which you have to prove).

To actually show it (using a set is open iff it equals its interior):

let $p \in V_g(c, \delta)$. Then $|p -c| < \delta$. Define $\epsilon = \delta - | p-c| > 0$. Then $V_g(p, \epsilon) \subseteq V_g(c, \delta)$: let $q \in V_g(p,\epsilon)$, then $|q -c | \le |q- p | + |p - c| < \epsilon + |p-c| = \delta$, so $q \in V_g(c, \delta)$. This shows that every point $p \in V_g(c,\delta)$ is an interior point of it, so $V_g(c,\delta)$ is an open set.