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

• I
• CGandC
In summary, the convex continuous function ##f:X\to \mathbb{R}## is continuous and convex on the interior of the closed ball around the point ##c##.
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?

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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
$$\mathrm{cl}\,\{x\in X\mid f(x)<c\}=\{x\in X\mid f(x)\le c\}$$

Last edited:
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
wrobel said:
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.

andrewkirk said:
@CGandC Yes your proof works.
Thanks for the help!

## 1. What is Euclidean space?

Euclidean space is a mathematical concept that describes a flat, infinite, and continuous space. It is named after the ancient Greek mathematician Euclid and is the basis for classical geometry.

## 2. What is a closed ball in Euclidean space?

A closed ball in Euclidean space is a set of all points within a given distance (radius) from a center point. It includes both the boundary points and the interior points of the ball.

## 3. What is an open ball in Euclidean space?

An open ball in Euclidean space is a set of all points within a given distance (radius) from a center point, excluding the boundary points. It only includes the interior points of the ball.

## 4. How is the closed ball related to the closure of an open ball in Euclidean space?

In Euclidean space, the closure of an open ball is equal to the closed ball. This means that the closure of an open ball contains all the points of the open ball, as well as its boundary points.

## 5. Why is the concept of closed ball and closure of open ball important in Euclidean space?

The concept of closed ball and closure of open ball is important in Euclidean space because it helps define important properties such as continuity and completeness. It also plays a crucial role in various mathematical proofs and applications in fields such as physics and engineering.

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