Prove that if F={x}⊂RnF=\{x\}\subset\mathbb{R}^n, there exist open sets U1,U2,…U_1,U_2,\ldots such that F=U1∩U2∩⋯F = U_1\cap U_2 \cap \cdots.

Thoughts so far: from the definition of any open set we know for any UiU_i there is a open ball of radius rr around every vector a∈Ua \in U (i.e.Br(a)⊆UB_r(a)\subseteq U). Secondly from the nested interval theorem the intersection of nested intervals is non-empty. I am not sure how to combine these two theorems.

=================

I’m not sure if I understand the question fully. Is FF an arbitrary closed set or is F={x}F=\{x\}? If F={x}F=\{x\} you can use Ui=B1/i(x)U_i=B_{1/i}(x) for instance.

– Olivier Moschetta

Oct 20 at 19:53

1

You don’t need any fancy theorems for this. Just consider balls with center xx with smaller and smaller radius.

– D_S

Oct 20 at 19:54

@OlivierMoschetta F is an arbitrary closed set with one element →x\vec{x}. Such that F={→x}F=\{\vec{x}\}. I will add this to the problem to make it clearer.

– AzJ

Oct 20 at 19:58

@AzJ: D_S has already answered your question. Also, F=xF=x is a typo.

– parsiad

Oct 20 at 20:00

1

Open ball centered in xx with radius r=1/ir=1/i.

– Olivier Moschetta

Oct 20 at 20:08

=================

1 Answer

1

=================

Let (X,d)(X,d) be a metric space, x∈Xx\in X, and F={x}F=\{x\}.

Remark: In your case, X=RnX=\mathbb{R}^n and d(x,y)=|x−y|d(x,y)=|x-y| is the absolute value distance between numbers. If you get confused by the use of dd below, simply replace instances of d(x,y)d(x,y) with |x−y||x-y|.

Definition: B(v;r)={w∈X:d(v,w)

2

Also relevant are the following facts: (i) a singleton set is always closed in a metric space (see math.stackexchange.com/questions/1459067/…) and (ii) Q\mathbb{Q} is an example of a closed set (in R\mathbb{R}) that cannot be written as a countable intersection of open sets. See also the definition of a GδG_\delta set: en.wikipedia.org/wiki/G%CE%B4_set.

– parsiad

Oct 20 at 20:17

That you for your answer but I believe you mean intersection instead of union (∩\cap instead of ∪ \cup).

– AzJ

Oct 20 at 20:19

Just one more question. What if we instead have a large number of vectors that made up FF (i.e. let FF be an infinite countable closed subset). What changes would we need to make to the proof. I think that instead of each open ball being around a single point in Rn\mathbb{R}^n there is a open ball around each point in the set FF. If you could comment on this change that would be great.

– AzJ

Oct 21 at 0:52

1

If FF is finite, this works. However, I have already given you (in the comment above) an example of a closed countably infinite set Q⊂R\mathbb{Q}\subset\mathbb{R} that cannot be written as an intersection of open sets (i.e., your claim is not true).

– parsiad

2 days ago

That makes sense thank you for the clarification.

– AzJ

2 days ago