Closure of half (open) ball included in the open ball - Proof verification

Question

Is $\overline{V_{\frac{r}{2}}(p)} \subset V_r (p)$ ?

My proof (by contradiction):

Clearly $${V_{\frac{r}{2}}(p)} \subset V_r (p)$$ So let's prove $LP( {V_{\frac{r}{2}}(p)} ) \subset V_r (p)$

In order to do so, assume it is not included. We then have $$LP( {V_{\frac{r}{2}}(p)} ) \cap X\backslash V_r (p) \neq \emptyset$$ Let's call $q$ a point belonging to this intersection.

$q$ being a LP of ${V_{\frac{r}{2}}(p)} $, we can take the open neighborhood ${V_{\frac{r}{2}}(q)}$ and have $${V_{\frac{r}{2}}(q)} \cap {V_{\frac{r}{2}}(p)} \backslash\{ q\}\neq \emptyset$$ On the other hand, $q$ belonging to $X\backslash V_r (p) $ implies that $|q-p| \geq r$.

Let's call $q'$ a point belonging to the last intersection we mentionned.

$q'$ belonging to ${V_{\frac{r}{2}}(q)}\backslash \{ q\} $ implies $|q'-q| < \frac{r}{2}$

$q'$ belonging to ${V_{\frac{r}{2}}(p)} \backslash\{ q\}$ implies $|q'-p|< \frac{r}{2}$

We shall now consider this quantity:

$$|q-p| \leq |q-q'| + |q'-p|$$ $$|q-p| < \frac{r}{2} + \frac{r}{2} $$ $$|q-p| < r $$

Contradiction with the previous line.

PS: $V_r(p)$ denotes the open ball (or open neighborhood) centered at $p$ and of radius $r$.

Questions:

Is there a way to prove the inclusion (faster?) without going through this negation of the inclusion?

Eventually, are there any imperfections to the proof from above?

if $x\in LP(V_{\frac{r}{2}}(p))$ then $|x-p|\le\frac{r}{2}<r$ so $x\in V_{r}(p)$ — janmarqz, Feb 08 '23 at 17:27
It is probably better to write out "limit point" on a public post. I don't think LP is a widely used abbreviation. — User6739481, Feb 08 '23 at 17:29
@janmarqz I think you are assuming that since $x\in LP$, $x$ is in the closure and hence is in the closed ball (of radius $\frac{r}{2}$). But this post shows it is not always true. Perhaps you're right, but it seems unclear to me — niobium, Feb 08 '23 at 17:54
any point with $|x-p|>\frac{r}{2}$ can't be limit point for $\overline{V_{\frac{r}{2}}(p)}$ , isn't it? — janmarqz, Feb 09 '23 at 01:47

score 1 · Answer 1 · answered Feb 08 '23 at 17:33

Possibly a faster way would be to prove that for every $x\in \overline{V_{r/2}(p)}$, one has $|x-p|\leq r/2$. Indeed, if $x\in \overline{V_{r/2}(p)}$, there exists a sequence $x_n\in V_{r/2}(p)$ such that $|x_n-x|\to 0$ as $n\to\infty$, hence $$|x-p|\leq |x-x_n|+|x_n-p|\leq |x_n-x|+r/2$$ and letting $n\to\infty$ we obtain $|x-p|\leq r/2$.

In particular, if $x\in \overline{V_{r/2}(p)}$, then $|x-p|<r$.

score 1 · Accepted Answer · answered Feb 08 '23 at 17:34

I think you can make a direct proof: Take $\varepsilon<r/3$. If $x$ is in the closure of the small ball, there is some $q\in V_{r/2}(p)$ within distance $\varepsilon$ from $x$. But all points within distance $\le r/3$ of a point in $V_{r/2}(p)$ belong to $V_r(p)$, so $x$ is in $V_r$.

Closure of half (open) ball included in the open ball - Proof verification

My proof (by contradiction):

Questions:

2 Answers2