Prove that a bounded a bounded set in $\mathbb{R^2}$ with a finite number of accumulation points has content 0.

Prove that a bounded set in $$\mathbb{R^2}$$ with a finite number of accumulation points has (Jordan ) content 0.

My thoughts:

First the definition of a set has (Jordan) content 0 given in the book is given below:

A set $$D \subset \mathbb{R^2}$$ has (Jordan) content 0 if for every $$\epsilon > 0$$ there exists a finite collection of rectangles $$R_{k},$$ $$1 \leq k \leq n,$$ whose union covers $$D$$ and the sum of their areas is less than $$\epsilon.$$

I know for example that the unit circle has Jordan content zero and also the unit sphere. my question is what is the importance of considering the accumulation points and why specifically they must be finite?

Any help on how to prove this will be appreciated, thanks!

• Cover the finite collection of accumulation points with an open set. What can you say about the complement of the open set intersected with the bounded set? – copper.hat Apr 15 at 18:54
• A closed bounded set @copper.hat – Secretly Apr 15 at 20:24
• Yes, but it is even more restrictive than that. – copper.hat Apr 15 at 20:30
• Could you explain your last comment a little bit more please?@copper.hat – Secretly Apr 15 at 21:38
• So your example is disqualified. – user58697 Apr 16 at 0:29

A single point has Jordan content zero. A finite collection of points has Jordan content zero.

Let $$C$$ be the set in question and let $$A$$ be the finite set of accumulation points.

Let $$U$$ be an open set such that $$A \subset U$$. Let $$F=C \setminus U$$.

I claim that $$F$$ is finite. Suppose not, then there are $$x_n \in F$$ such that the $$x_n$$ are distinct and countable. Since $$C$$ is bounded, there is some $$x$$ and a subsequence such that $$x_{n_k} \to x$$. Since every neightbourhood of $$x$$ contains some point of $$F$$, we see that $$x \in A$$, which is a contradiction. Hence $$F$$ is finite.

So, pick $$\epsilon>0$$ and choose rectangles of total area $$< {1 \over 2} \epsilon$$ that contain the accumulation points in their interior. Let $$U$$ be the interior of the union of rectangles then $$C \setminus U$$ is finite, and we can find rectangles that cover these with total area $$< {1 \over 2} \epsilon$$.

• do you mean $x_{n}$ countably infinite ? – Secretly Apr 16 at 6:16

You have, say $$N$$ accumulation points. Given $$\epsilon$$, cover each of them with the rectangles of an area $$\dfrac{\epsilon}{2N}$$ each, $$\dfrac{\epsilon}{2}$$ total. You still have a remaining budget of $$\dfrac{\epsilon}{2}$$ to cover the points yet uncovered.

The only thing left to prove is that there is finite amount of such points. Hint: there is no more accumulation points.

• you mean after covering the accumulation points , I will still have remaining finite number of points that is uncovered? – Secretly Apr 16 at 0:44
• just to really nail down the idea .... could you provide the details please? – Secretly Apr 16 at 2:10
• I think I will still have remaining finite number of points that is uncovered because the given set is bounded ...... correct? – Secretly Apr 16 at 2:17