# In Euclidian space, closed ball is equal to closure of open ball

• I
• CGandC

#### CGandC

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 ].

How to prove the "## ( \supseteq ) ##" side? I thought maybe I'd use the theorem "## x \in cl(B) \iff ## there exists a sequence ## x_n \in B ## , ## x_n \rightarrow x ## "; that means I'd show the existence of a sequence ## (x_k)_{k=1} \subseteq R^n ## s.t. ## (x_k)_{k=1} = ((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 ## , but the question is how to define this sequence of sequences?

You don't need a sequence of sequences.
For a point ##x## on the boundary of the closed ball, why not just approach it by the sequence that goes along the line segment from ##c## to ##x##, starting at ##c## and at each step jumping to halfway between current location and ##x##. You should be able to prove that converges to ##x## and that all its points are in the open ball.

Ok, here's what I've did:
I started drawing an intuitive ( non-precise ) picture of what you've said, I've noted that In the attempted proof I gave I had mistaken with the arbitrary ## r ## I introduced with the ## r ## given in the question. Here's my attempted proof based on your idea:

Let ## \tau \in cl(B(c,r)) ## be arbitrary. Hence ##d(c,\tau) \leq r##.
Take the sequence ## (x_m) \subseteq \mathbb{R}^n ## defined as follows,
For every ## m \in \mathbb{N} ##, ## x_m = c + \sum_{k=1}^{m}\frac{\tau-c}{2^i} = c + \frac{1}{2}\cdot(\tau-c)\sum_{k=0}^{m-1} \frac{1}{2^k} = c + ( \tau - c)\cdot(1-(\frac{1}{2})^m) ##
Note that ## \lim_{m\to\infty}x_m = \tau ##.
Also note that ## x_m ## is a vector equation, meaning that for all ## 1 \leq i \leq n ## , ## (x_m)_i = c_i + ( \tau_i - c_i)\cdot(1-(\frac{1}{2})^m) ##, from this, note that ## (x_m) \subseteq B(c,r) ## since for all ## m \in \mathbb{N} ## we have that ## d(c,x_m) = \sqrt{ \Bigg( c_1 - \bigg( c_1 + ( \tau_1 - c_1)\cdot(1-(\frac{1}{2})^m) \bigg) \Bigg)^2 + \cdots + \Bigg( c_n - \bigg( c_n + ( \tau_n - c_n)\cdot(1-(\frac{1}{2})^m) \bigg) \Bigg)^2 } =
( 1 - (\frac{1}{2})^m ) \sqrt{ ( \tau_1 - c_1)^2 + \cdots + ( \tau_n - c_n)^2 }.## Since ## d(c,\tau) \leq r ## we can see that ## d(c,x_m) < r ##, hence ## (x_m) \subseteq B(c,r) ##. So since ## \lim_{m\to\infty}x_m = \tau ## and ## \tau ## was arbitrary, we're finished.

Do you think this is ok?

#### Attachments

Sometimes a general fact is much easier to prove because the key points are presented explicitly.

Let ##f:X\to \mathbb{R}## be a convex continuous function on the normed vector space ##X##. Then
$$\mathrm{cl}\,\{x\in X\mid f(x)<c\}=\{x\in X\mid f(x)\le c\}$$

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• sysprog
Thanks but I haven't learned about convex functions and normed vector spaces yet (I'm doing a real analysis course and just very little of the basics of topology are taught). It seems to me that If I can create such a function as you've said, the proof will be finished.

Convex function is a function such that the inequality
$$f(\lambda x_1+(1-\lambda)x_2)\le \lambda f(x_1)+(1-\lambda)f(x_2)$$
holds for all $$x_1,x_2\in X,\quad \lambda\in[0,1].$$
Take ##f(x)=\|\tilde x-x\|## and ##X=\mathbb{R}^m##

• sysprog
Convex function is a function such that the inequality
$$f(\lambda x_1+(1-\lambda)x_2)\le \lambda f(x_1)+(1-\lambda)f(x_2)$$
holds for all $$x_1,x_2\in X,\quad \lambda\in[0,1].$$
Take ##f(x)=\|\tilde x-x\|## and ##X=\mathbb{R}^m##

Ok now I understand how the general theorem you proposed relates to the problem. In the private case you proposed I get what I'm asked to prove. How do I prove the "## \supseteq ##" direction? got a hint please?

Here's what I've done ( had no clue how to proceed beyond the beginning ):
Let ## x ## be arbitrary such that ## f(x) \leq c ##. Let ## r>0 ##. We'll show that ## B(x,r) \subseteq \{ y \in X : f(y) < c \} ##. Let ## \tau \in B(x,r) ##, thus ## d(x,\tau)<r ##. [ I don't see how can I relate ## \tau ## to ## f ## in order to have ## f(\tau) < c ##. ]

I then thought maybe I'd find a sequence ## x_n \in \{ y \in X : f(y) < c \} ## which converges to ## x ## and that'd show ## x ## is in ## \mathrm{cl}\,\{x\in X\mid f(x)<c\} ## , but couldn't think of anything that'd help me allow to do that.