Finite Cover Length Estimates by Ordered Endpoints

Why Order the Endpoints?

A finite collection of intervals can overlap, nest inside one another, or meet at endpoints. Those features make it tempting to reason from a sketch of the cover, but a sketch does not establish a length bound. Ordering the endpoints gives a finite list of elementary gaps. Each gap can be traced to an interval that covers it, and the gaps assigned to any one interval cannot have greater total length than that interval. This provides a direct proof of the finite-cover length estimate without assuming that the intervals are disjoint or arranged from left to right.

All intervals below are subsets of $\mathbb{R}$. For a bounded interval with infimum $u$ and supremum $v$, define its length to be $v-u$, regardless of whether either endpoint belongs to the interval. Thus a singleton has length zero. An unbounded interval has length $+\infty$. We consider a target interval with finite endpoints $a\le b$; when $a<b$, the target contains every point strictly between $a$ and $b$, whatever its endpoint inclusion convention. The case $a=b$ is a singleton.

Definition: A finite interval cover of a target interval $J$ is a finite list of intervals $I_1,\ldots,I_N$ such that $J\subseteq\bigcup_{j=1}^N I_j$. Its total length is $\sum_{j=1}^N\ell(I_j)$, where $\ell(I_j)$ is the interval length just defined. Repeated intervals are allowed and are counted each time they occur in the list.

The Endpoint-Gap Argument

Suppose first that $a<b$ and that all covering intervals are bounded. Form a finite set from $a,b$ and every endpoint of every $I_j$ that lies in $[a,b]$. Remove duplicates and list the resulting numbers in increasing order:

a=x_0<x_1<\cdots<x_r=b.

The open intervals $(x_{k-1},x_k)$, for $1\le k\le r$, are the elementary gaps. Their lengths add to the target's length, since the sum telescopes:

\sum_{k=1}^{r}(x_k-x_{k-1})=b-a.

Each gap has a covering interval that contains the whole gap. To see this, choose its midpoint $z_k=(x_{k-1}+x_k)/2$. Since $z_k\in(a,b)$, it belongs to the target and hence to at least one covering interval. Select one such interval. No endpoint of any covering interval lies strictly inside $(x_{k-1},x_k)$, by the way the list was formed. An interval containing $z_k$ therefore cannot start or stop between $z_k$ and another point of this gap. The defining property of an interval then gives $(x_{k-1},x_k)\subseteq I_j$ for the selected interval $I_j$. Endpoint inclusion does not affect this assertion about the open gap.

Lemma (Ordered-gap estimate): Let $I_1,\ldots,I_N$ be bounded intervals, and let $a=x_0<\cdots<x_r=b$ be an ordered partition whose points include all their endpoints in $[a,b]$. If each elementary gap $(x_{k-1},x_k)$ is assigned to an interval containing it, then the sum of the lengths of the gaps assigned to $I_j$ is at most $\ell(I_j)$, for each $j$.

Proof. Write $I_j$'s infimum and supremum as $u_j$ and $v_j$, so that $\ell(I_j)=v_j-u_j$. List the gaps assigned to $I_j$ from left to right, with endpoints $(\alpha_1,\beta_1),\ldots,(\alpha_s,\beta_s)$. Each open gap lies inside $I_j$, and consequently $u_j\le\alpha_q<\beta_q\le v_j$ for each $q$. Distinct gaps have disjoint interiors, so $\beta_q\le\alpha_{q+1}$ for $1\le q<s$. If $s\ge1$, then \[ \sum_{q=1}^{s}(\beta_q-\alpha_q) =\beta_s-\alpha_1-\sum_{q=1}^{s-1}(\alpha_{q+1}-\beta_q) \le \beta_s-\alpha_1 \le v_j-u_j. \] If $s=0$, the sum is zero and is also at most $v_j-u_j$. This proves the estimate, including when $I_j$ is a singleton. $\square$

Theorem (Finite-cover length bound): If a finite list of intervals $I_1,\ldots,I_N$ covers a target interval $J$ with finite endpoints $a\le b$, then $$ b-a\le \sum_{j=1}^N\ell(I_j). $$ The intervals may be open, closed, half-open, or degenerate. If any covering interval is unbounded, the right-hand side is $+\infty$, and the inequality holds in the extended nonnegative real numbers.

Proof. If $a=b$, the left-hand side is zero and every interval length on the right is nonnegative, so the claim holds. If a covering interval is unbounded, its length is $+\infty$, as noted in the theorem, and the inequality follows. It remains to consider $a<b$ with all covering intervals bounded. Construct the ordered endpoint partition and assign each gap to a covering interval as described above. Each gap is assigned exactly once, so the sum of the lengths of all gaps is the sum, over $j$, of the lengths of the gaps assigned to $I_j$. The lemma bounds each of these latter sums by $\ell(I_j)$. Therefore \[ b-a=\sum_{k=1}^{r}(x_k-x_{k-1}) \le\sum_{j=1}^N\ell(I_j), \] as required. $\square$

Working Through Finite Covers

Worked Example: Overlapping Intervals with Endpoints Inside the Target

Consider the cover of $[0,6]$ by $I_1=[-1,5/2]$, $I_2=(2,4]$, and $I_3=[7/2,7]$. Verify the cover and use ordered endpoints to account for the target's length.

The interval lengths are \[ \ell(I_1)=5/2-(-1)=7/2,\qquad \ell(I_2)=4-2=2,\qquad \ell(I_3)=7-7/2=7/2. \] Their total is $7/2+2+7/2=9$. Within the target, the ordered endpoints are $0,2,5/2,7/2,4,6$. The five gap lengths are, in order, \[ 2-0=2,\quad 5/2-2=1/2,\quad 7/2-5/2=1,\quad 4-7/2=1/2,\quad 6-4=2. \] They sum to $2+1/2+1+1/2+2=6$. Assign the first two gaps to $I_1$, the middle gap to $I_2$, and the final two gaps to $I_3$. The assigned totals are $2+1/2=5/2\le7/2$, $1\le2$, and $1/2+2=5/2\le7/2$, respectively. These assignments show the entire interior is covered. The endpoints $0,2,5/2,7/2,4,6$ are covered as well: $0,2\in I_1$, $5/2\in I_1$, $7/2\in I_2\cap I_3$, $4\in I_2\cap I_3$, and $6\in I_3$. Check: the target length is $6-0=6$, while the total covering length is $9$, so $6\le9$.

Worked Example: Endpoint Inclusion in a Cover

Show directly that $[1,4]$ is covered by $[0,2)$, $[2,3]$, and $(3,5]$, and apply the estimate.

The interval lengths are $2-0=2$, $3-2=1$, and $5-3=2$; the total is $2+1+2=5$. The ordered points in the target are $1,2,3,4$, giving gaps $(1,2)$, $(2,3)$, and $(3,4)$, each of length $1$. The first gap lies in $[0,2)$, the second in $[2,3]$, and the third in $(3,5]$. The target endpoints are covered: $1\in[0,2)$ and $4\in(3,5]$; the shared boundary points $2$ and $3$ are in $[2,3]$. Thus this is indeed a cover, and the estimate reads $4-1=3\le5$. Check: the three assigned gaps total $1+1+1=3$, exactly the target length, and each assigned length $1$ is at most its assigned interval's length $2,1,2$, respectively.

Worked Example: A Cover Attaining Equality

Consider $[-2,3]$ covered by $[-2,0)$, $[0,1]$, and $(1,3]$. Compute both sides of the bound and check endpoint coverage.

The three lengths are $0-(-2)=2$, $1-0=1$, and $3-1=2$, so their sum is $2+1+2=5$. The ordered endpoints $-2,0,1,3$ give gaps of lengths $0-(-2)=2$, $1-0=1$, and $3-1=2$. Each gap lies in the corresponding interval. The endpoints are covered: $-2$ belongs to the first interval, $0$ and $1$ belong to the second, and $3$ belongs to the third. Hence the full target is covered, and its length is $3-(-2)=5$. Check: the target length and the total covering length are both $5$, so equality holds.

Endpoint Errors and the Scope of the Estimate

A frequent mistake is to treat two adjacent open intervals as if they covered the interval obtained by joining them. For example, $(0,1)$ and $(1,2)$ have total length $1+1=2$, but they do not cover $[0,2]$: neither interval contains $0$, $1$, or $2$. The calculation $2\ge2$ does not repair the missing points; the estimate applies only after a genuine cover has been established.

For a corrected cover, use $[0,1]$ and $(1,2]$. These intervals cover $[0,2]$: the first contains $0$ and $1$, and the second contains every point strictly greater than $1$ up to and including $2$. Their lengths are $1-0=1$ and $2-1=1$, so the total is $2$, equal to the target length. A singleton interval $\{1\}$, whose length is zero, could also be added to the original two open intervals to cover the missing interior point, but it would not supply the missing endpoints $0$ and $2$.

The estimate is a lower bound on the total length, not a claim that the intervals form a disjoint partition or that their union has length equal to the sum. Overlaps are allowed and may make the sum strictly larger than the target length. Nor does the finite argument require the covering intervals to be ordered by their left endpoints with their right endpoints in the same order; nested intervals show why such an assumption would be unjustified. The endpoint partition instead orders every boundary point and handles each elementary gap explicitly.

This finite estimate is useful whenever a cover can be reduced to finitely many intervals and a lower bound on their total length is needed. Its proof depends on finiteness because it creates a finite endpoint list and assigns each resulting gap. It does not, by itself, establish a length bound for a countable cover by repeating the same finite partition argument; countable-cover arguments require their own justification.

Check Your Understanding

For each question, distinguish the coverage condition from the numerical length calculation.

Let $[0,5]$ be covered by finitely many bounded intervals. What points should be included in the ordered endpoint list?
Why does choosing a midpoint in an elementary gap identify an interval containing the whole open gap?
Can a finite cover of $[0,3]$ have total interval length $5/2$? Explain using the theorem.
Do $(0,1)$ and $(1,2)$ cover $[0,2]$? Which points are missing, and what is their total length?
What does the theorem say when one of the covering intervals is unbounded?

Finite Cover Length Estimates by Ordered Endpoints

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