Real Analysis – Proof that Interval [0,1] Cannot Have Zero Content

calculusreal-analysis

Using the following definition of zero content, I wish to prove that the interval $[0,1]$ does not have zero content.

A set $A$ has zero content if $\forall \epsilon > 0 \exists$ finite collection of closed intervals $I_1, \cdots, I_L$ such that $A \subset \cup_{l=1}^L I_l$ and $\sum_{l=1}^L length(I_l) < \epsilon$.

Intuitively, it is clear that any set of closed intervals covering $[0,1]$ must have total length of at least 1. To formally prove this, I assumed that $I_1, \cdots, I_L$ covers $[0,1]$, but I am not sure how to proceed.

Your help is much appreciated!

Best Answer

Assume that $[0, 1]$ is covered by the intervals. Discard all the intervals that don't intersect $[0, 1]$, and also discard all the intervals that are subsets of other intervals. The remaining intervals, which I'll call $J_1,\dots,J_N$ will still cover $[0, 1]$. Let the intervals $J_n = [a_n, b_n]$ be ordered such that the sequence of $b_n$'s is increasing.

It must be the case that $b_n \geq a_{n+1}$ whenever $1 \leq n \leq N - 1$, which we can prove by contradiction. If $b_n < a_{n + 1}$, then consider some $x$ between $b_n$ and $a_{n + 1}$. Since $b_n < 1$ and $a_{n+1} > 0$, we have $x \in [0, 1]$. Therefore $x$ must be contained in $J_k$ for some $k$.

  • If $k \leq n$, then $b_k \leq b_n < x$, which is impossible since $x \in J_k$.
  • If $k > n + 1$, then $a_k \leq x < a_{n + 1}$. But since $b_k > b_{n + 1}$, it must be that $J_{n+1} \subset J_k$, which is impossible since we excluded all intervals which were subsets of other intervals.
  • $k$ cannot equal $n + 1$, since $x < a_{n + 1}$.

We've found that every possible value of $k$ leads to a contradiction, so it must be that $b_n \geq a_{n + 1}$.

Now we can apply this inequality to prove the result: \begin{align} \sum_{n = 1}^L \text{length}(I_n) &\geq \sum_{n = 1}^N \text{length}(J_n)\\ &= \sum_{n = 1}^N (b_n - a_n)\\ &\geq \sum_{n = 1}^{N - 1} (a_{n+1} - a_n) + (b_N - a_N)\\ &= b_N\\ &\geq 1. \end{align}

I am not sure where your comfort level is, but I skipped a few justifications that I felt weren't worth the amount of space they would take to write. In particular, I did not show that the $J$ intervals really do cover $[0, 1]$, that $b_n < 1$ when $n \leq N - 1$, or that $a_{n+1} > 0$ when $n \geq 1$.

Related Question