Show that $M:= \{x\in \mathbb{R}^n~|~ \Vert x-a\Vert \leq \delta\}$ is closed

Question

Let's consider the following problem:

We want to show that $M:= \{x\in \mathbb{R}^n~|~ \Vert x-a\Vert \leq \delta\}\neq\emptyset$ is closed. According to our definition:

A set $M$ is closed iff $M$ contains all its limit points.

Let be $b$ a limit point of $M$. Then we find a point $x\in M$ with $x \neq b$ in every neigbourhood. So for every $\epsilon>0$ the open ball $U_{\epsilon}(b):=\{x\in M~|~ \Vert x-b\Vert < \epsilon\}$ is non empty and contains a point $x\neq b$. Then we let $\epsilon \to 0$ and simply conclude: $\Vert b-a\Vert = \Vert b-x+x-a\Vert \leq \Vert b-x \Vert+\Vert x-a\Vert\leq 0 + \delta$ which shows that $b\in M$. Hence, $M$ is closed as $b$ was arbitrarily chosen.

Actually this seems to me like a typical analysis technique but I am not sure if I have a wrong intuition in my mind or maybe the reasoning above is wrong?

In my mind $\epsilon$ will never reach $0$ it will only be very very close to it so that $\Vert b-x \Vert$ also will never become $0$. So following this logic the inequality above must be wrong?! May be someone can help me with this confusion.

Answer 1

I don't agree that we can say “we let $\varepsilon\to0$”. I would do that proof as follows: take $\varepsilon>0$. There is some $x\in M$ such that $\|b-x_\varepsilon\|<\varepsilon$. So,$$\|b-a\|\leqslant\|b-x_\varepsilon\|+\|x_\varepsilon-a\|<\delta+\varepsilon.$$Since this occurs for each $\varepsilon>0$, then$$\|b-a\|\leqslant\lim_{\varepsilon\to0}\delta+\varepsilon=\delta.$$Note that at no point I took $\varepsilon=0$.

Answer 2

Your argument $\Vert b-x \Vert+\Vert x-a\Vert\leq 0 + \delta$ is inadmissible because $b \ne x$.

Assume $\Vert b-a\Vert >\delta$. Then take $\epsilon = \Vert b-a\Vert -\delta > 0$ and conclude $$\Vert b-a\Vert = \Vert b-x+x-a\Vert \leq \Vert b-x \Vert+\Vert x-a\Vert < \epsilon + \delta = \Vert b-a\Vert$$ which is a contradiction.

Answer 3

We have $||b-a||\le||b-x||+||x-a||\lt\epsilon+\delta$ for all $\epsilon>0$.

I claim then $||b-a||\le\delta$,

because otherwise we'd have $||b-a||>\delta$, and then $\epsilon=||b-a||-\delta>0$, and

then $||b-a||=\epsilon+\delta$, so it would not be true that $||b-a||\lt\epsilon+\delta$ for all $\epsilon>0$.