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Problem: Let $ (X,d) $ be a metric space, denote as $ B(c,r) = \{ x \in X : d(c,x) < r \} $ the open ball at radius $ r>0 $ around $ c \in X $, denote as $ \bar{B}(c, r) = \{ x \in X : d(c,x) \leq r \} $ the closed ball and for all $ A \subset X $ we'll denote as $ cl(A) $ the closure of $ A $ ( sometimes denoted also as $ \bar{A} $ ).

Show that in $ \mathbb{R}^n $ with the standard metric it occurs that: $cl(B(c, r))=\bar{B}(c, r)$

Attempt:
$ ( \subseteq ) $ Let $ \tau \in cl(B(c,r)) $. There exists a sequence $ x_n \in B(c,r) $ s.t. $ x_n \rightarrow \tau $
( From a theorem that says: $ x \in cl(B) \iff $ there exists a sequence $ x_n \in B $ , $ x_n \rightarrow x $ ).
Note that for all $ n \in \mathbb{N} $, from the fact $ x_n \in B(c,r) $ we have $ d(c,x_n) < r $. Also, since $ x_n \rightarrow \tau $ we have that $ d(x_n,\tau) \rightarrow 0 $.
So by triangle inequality, we have for all $ n \in \mathbb{N} $ that $ d(c,x) \leq d(c,x_n) + d(x,x_n) $, taking the limit we get $ d(c,x) \leq r $.

$ ( \supseteq ) $ Let $ \tau \in \bar{B}(c,r) $, hence $ d(c,\tau) \leq r $.
( Now we want to show that $ \tau \in cl(B(c,r)) $, meaning for all $ r>0$ we want to show $ B(\tau,r) \cap B(c,r) \neq \emptyset $ )
Let $ r>0 $. Notice that $ \tau \in B(\tau,r) $ since $ d(\tau,\tau) =0 < r $. In addition we have $ \tau \in \bar{B}(c,r) $ then $ d(\tau,c) \leq r $.
If $ d(\tau,c)<r $ then $ \tau \in B(c,r) $, hence $ \tau \in B(\tau,r) \cap B(c,r) $.
If $ d(\tau,c) = r $ then [ missing arguments for completing proof ].

Questions:

  1. How to continue in the "$ ( \supseteq ) $" part to finish the proof?
  2. I don't see where I have used the fact that our metric space is $ \mathbb{R^n} $ and the metric is the standard metric in $ \mathbb{R^n} $, do you have any Idea?
  3. Is my proof for the " $ ( \subseteq ) $ " part ( which doesn't use the fact that the metric space is $ \mathbb{R^n} $ and the metric is the standard metric in $ \mathbb{R^n} $ ) is a proof for the following statement: "In every metric space, $ cl(B(c, r)) \subseteq \bar{B}(c, r) $" ?
hazelnut_116
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    Check also this: https://math.stackexchange.com/q/108010/42969 and these: https://math.stackexchange.com/questions/linked/108010. – Martin R Feb 10 '22 at 09:16

1 Answers1

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$ cl(B(c, r)) \subseteq \bar{B}(c, r) $ is true in any metric space. For the reverse inclusion in the case of the usual metric on $\mathbb R^{n}$ let $d(x,c) \leq r$. Let $x_k=c+(1-\frac 1 k) (x-c)$. Verify that $d(x_k,c ) <r$ for each $k$ and $x_k \to x$ as $k \to \infty$. This shows that $x$ belongs to the closure of $B(c,r)$.

Kavi Rama Murthy
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  • Just to make sure, $ (x_k){k=1} \subseteq R^n $ is supposed to be a sequence $ ((x^{(1)}_i){i=1}^n,(x^{(2)}i){i=1}^n,... ) $ s.t. $ x_k = (x^{(k)}i){i=1}^n $ s.t. $ (x^{(k)}i){i=1}^n \in R^n $, and you've defined $ x^{(k)}_i $ as $ x^{(k)}_i = c+(1-\frac 1 k) (x-c) $? – hazelnut_116 Feb 10 '22 at 12:05
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    @hazelnut_116 $x_n=c+(1-\frac 1 n) (x-c)$ is vector equation. It means $(x_n)_i=c_i+(1-\frac 1 n) (x_i-c_i)$ for $1 \leq i \leq n$. – Kavi Rama Murthy Feb 10 '22 at 12:47