Half-Open Intervals

One Included Endpoint and One Excluded Endpoint

An interval can include one endpoint and exclude the other. These are the two half-open forms: $[a,b)$, which includes $a$ but not $b$, and $(a,b]$, which excludes $a$ but includes $b$. They are called half-open because each has one included and one excluded endpoint; “half-closed” is another name for the same intervals.

In each case, assume $a<b$. The order of the brackets matters: the square bracket marks an endpoint that belongs to the interval, while the parenthesis marks one that does not. The strict and non-strict inequalities in the set descriptions make this distinction precise.

Definition (Half-open intervals). For real numbers $a<b$, the half-open interval $[a,b)$ is $\{x\in\mathbb{R}:a\leq x<b\}$, and the half-open interval $(a,b]$ is $\{x\in\mathbb{R}:a<x\leq b\}$. Thus $[a,b)$ includes $a$ and excludes $b$, while $(a,b]$ excludes $a$ and includes $b$.

Both sets are intervals in the between-points sense introduced earlier in the course. For example, if $x,z\in[a,b)$ and $x<y<z$, then $a\leq x<y<z<b$, so $y\in[a,b)$. For $(a,b]$, if $x,z\in(a,b]$ and $x<y<z$, then $a<x<y<z\leq b$, so $y\in(a,b]$. The included or excluded status of a boundary point does not disrupt the between-points property.

The definitions also settle membership at the endpoints directly. The point $a$ belongs to $[a,b)$, but not to $(a,b]$. The point $b$ belongs to $(a,b]$, but not to $[a,b)$. Every point strictly between $a$ and $b$ belongs to both.

Worked Example: Checking Endpoint Membership

Consider $[-4,3)$ and $(-4,3]$. For the first interval, $x\in[-4,3)$ means $-4\leq x<3$. Substituting $x=-4$ gives $-4\leq-4<3$, which is true, so $-4$ belongs. Substituting $x=3$ gives $-4\leq3<3$, which is false, so $3$ does not belong.

For the second interval, $x\in(-4,3]$ means $-4<x\leq3$. Substituting $x=-4$ fails the strict inequality $-4<-4$, while substituting $x=3$ gives $-4<3\leq3$, which is true. Thus $-4\notin(-4,3]$ and $3\in(-4,3]$. The point $0$ belongs to both because $-4<0<3$.

Comparing Half-Open Intervals

The inequalities describing an interval make it possible to test whether one half-open interval lies inside another. For intervals of the form $[a,b)$, the left endpoint of the smaller interval cannot lie to the left of the larger interval’s left endpoint. Its excluded right endpoint cannot extend past the larger interval’s excluded right endpoint. The next theorem makes both requirements exact.

Theorem (Inclusion criterion for half-open intervals). Suppose $a<b$ and $c<d$. Then $$ [a,b)\subseteq[c,d) \quad\Longleftrightarrow\quad c\leq a\ \text{and}\ b\leq d. $$

Proof. First suppose $c\leq a$ and $b\leq d$. Let $x\in[a,b)$. Then $a\leq x<b$, so

c\leq a\leq x<b\leq d.

In particular, $c\leq x<d$, and therefore $x\in[c,d)$. This proves $[a,b)\subseteq[c,d)$.

Conversely, suppose $[a,b)\subseteq[c,d)$. Since $a<b$, the point $a$ belongs to $[a,b)$, and hence belongs to $[c,d)$. It follows that $c\leq a<d$, so $c\leq a$. We now show that $b\leq d$. If instead $b>d$, then $a<b$ and $d<b$. Let $m=\max\{a,d\}$, so $m<b$, and set $x=(m+b)/2$. Since $m<b$, we have $m<x<b$; also $x\geq m\geq a$ and $x>m\geq d$. Thus $x\in[a,b)$ but $x\notin[c,d)$, contradicting the assumed inclusion. Therefore $b\leq d$. Both endpoint conditions are necessary, proving the equivalence. $\square$

Notice why the theorem compares the boundary numbers themselves, including the excluded right endpoints. Although $b\notin[a,b)$, the interval contains points arbitrarily close to $b$ from the left. If $b>d$, some of those points lie to the right of $d$, so they cannot all be contained in $[c,d)$.

Worked Example: Testing Inclusion

Determine whether $[2,7)\subseteq[1,8)$. Here the endpoints satisfy $1\leq2$ and $7\leq8$, so the inclusion criterion gives $[2,7)\subseteq[1,8)$. Directly, every $x\in[2,7)$ satisfies $1\leq2\leq x<7\leq8$, and hence belongs to $[1,8)$.

Now compare $[2,7)$ with $[1,6)$. The left endpoint condition $1\leq2$ holds, but the right endpoint condition $7\leq6$ fails. Indeed, $13/2$ belongs to $[2,7)$, since $2\leq13/2<7$, but it does not belong to $[1,6)$, since $13/2>6$. Therefore $[2,7)$ is not a subset of $[1,6)$.

Splitting an Interval Without Overlap

Half-open intervals are especially useful when one interval is divided into adjacent pieces. Suppose a cut point $c$ lies between $a$ and $b$. If the first piece includes $c$, then the second piece must exclude it to avoid overlap; if the first piece excludes $c$, then the second can include it. With the convention $[u,u)=\varnothing$, the following result also covers a cut made at either endpoint.

Theorem (Adjacent splitting of a half-open interval). If $a\leq c\leq b$, then $$ [a,b)=[a,c)\cup[c,b), $$ and the two intervals on the right are disjoint.

Proof. Let $x\in[a,b)$. Then $a\leq x<b$. By trichotomy, either $x<c$ or $c\leq x$. In the first case, $a\leq x<c$, so $x\in[a,c)$. In the second case, $c\leq x<b$, so $x\in[c,b)$. Thus every point of $[a,b)$ belongs to the union.

For the reverse inclusion, if $x\in[a,c)$, then $a\leq x<c\leq b$, so $x\in[a,b)$. If $x\in[c,b)$, then $a\leq c\leq x<b$, so again $x\in[a,b)$. Therefore the union equals $[a,b)$.

To prove disjointness, suppose a point $x$ belonged to both pieces. Membership in $[a,c)$ would give $x<c$, while membership in $[c,b)$ would give $c\leq x$. These inequalities cannot both hold. Hence the pieces are disjoint.

If $c=a$, then $[a,c)=[a,a)=\varnothing$, and the second piece is $[a,b)$. If $c=b$, then the second piece is $[b,b)=\varnothing$, and the first piece is $[a,b)$. These endpoint cases are included in the same proof. $\square$

Worked Example: Splitting at Two Cut Points

Split $[-4,8)$ at $-1$ and $3$. Applying the adjacent splitting theorem first at $-1$, and then splitting $[-1,8)$ at $3$, gives

[-4,8)=[-4,-1)\cup[-1,3)\cup[3,8).

These pieces cover the original interval: a point below $-1$ but at least $-4$ lies in the first piece; a point at least $-1$ but below $3$ lies in the second; and a point at least $3$ but below $8$ lies in the third. The cut points are assigned exactly once: $-1$ belongs to $[-1,3)$, not $[-4,-1)$, and $3$ belongs to $[3,8)$, not $[-1,3)$. The endpoint $-4$ is included, while $8$ is excluded. Thus the pieces have no gaps and no overlaps.

Endpoint Notation Is Not Set Openness

A half-open interval is an interval, but the word “half-open” describes its endpoint inclusion, not a guarantee that the set is open or closed in the neighborhood sense. For $a<b$, the previous tutorial proved that $[a,b)$ is neither open nor closed. The same is true of $(a,b]$, as the following argument verifies for the other endpoint convention.

Proposition (Neither half-open form is open or closed). If $a<b$, then both $[a,b)$ and $(a,b]$ are neither open nor closed subsets of $\mathbb{R}$.

Proof. The claim for $[a,b)$ is the proposition established in the previous tutorial. Consider $(a,b]$. The point $b$ belongs to this set. For any $\varepsilon>0$, let $y=b+\varepsilon/2$. Then $|y-b|=\varepsilon/2<\varepsilon$, but $y>b$, so $y\notin(a,b]$. Thus no neighborhood of $b$ is contained in $(a,b]$, and this interval is not open.

To show it is not closed, consider its complement and the point $a$, which lies in the complement because $a$ is excluded. Given any $\varepsilon>0$, define $\delta=\frac12\min\{\varepsilon,b-a\}$, so $0<\delta<\varepsilon$ and $\delta\leq(b-a)/2$. Set $y=a+\delta$. Then $y>a$, and

y=a+\delta\leq a+\frac{b-a}{2}<b.

Consequently $y\in(a,b]$, while $|y-a|=\delta<\varepsilon$. Every neighborhood of $a$ therefore meets $(a,b]$, so no neighborhood of $a$ lies entirely in its complement. The complement is not open, and hence $(a,b]$ is not closed. This proves both claims. $\square$

This distinction prevents a common confusion. A square bracket means “included endpoint,” not “closed set”; a parenthesis means “excluded endpoint,” not “open set.” For nondegenerate half-open intervals, each endpoint convention gives an interval that is neither open nor closed as a subset of $\mathbb{R}$. Membership and openness or closedness are different questions and should be checked using their respective definitions.

Worked Example: Checking the Other Half-Open Form

Consider $(2,9]$. Its included endpoint $9$ prevents openness: for any $\varepsilon>0$, the point $9+\varepsilon/2$ is within distance $\varepsilon$ of $9$, but is outside the interval. Its excluded endpoint $2$ prevents closedness: choose $\delta=\frac12\min\{\varepsilon,7\}$ and take $y=2+\delta$. Then $0<\delta\leq7/2$, so $2<y<9$, and therefore $y\in(2,9]$, while $|y-2|=\delta<\varepsilon$. Every neighborhood of the complement point $2$ meets the interval, so the complement is not open.

Why the Half-Open Convention Is Useful

The splitting theorem explains a practical advantage of half-open intervals: adjacent pieces can cover a full interval while remaining disjoint. If both pieces included their shared endpoint, that point would be counted twice. If both excluded it, the point would be missing. Assigning a cut point to exactly one piece gives a clean decomposition.

The direction of the brackets must still match the intended assignment. The formula $[a,c)\cup[c,b)$ assigns the cut point $c$ to the piece on its right. Alternatively, $[a,c]\cup(c,b)$ assigns it to the piece on its left; that version also covers $[a,b)$ when $a\leq c<b$, with the endpoint cases considered separately. These conventions are useful in organizing intervals into nonoverlapping pieces, but they do not change which points are real numbers or alter the order properties of the real line.

Key takeaway. For $a<b$, $[a,b)$ includes its left endpoint and excludes its right, while $(a,b]$ does the reverse. Half-open intervals can be compared by their endpoint inequalities and split into disjoint adjacent pieces by assigning each cut point to exactly one piece. Their endpoint notation does not make them open or closed sets.

Check Your Understanding

Use the inequalities in the definitions and the results proved above to answer each question.

Which of the two intervals $[a,b)$ and $(a,b]$ contains $a$, and which contains $b$, when $a<b$?
For $a<b$ and $c<d$, what two endpoint inequalities are equivalent to $[a,b)\subseteq[c,d)$?
Why are $[a,c)$ and $[c,b)$ disjoint when $a\leq c\leq b$?
In the decomposition $[a,b)=[a,c)\cup[c,b)$, which piece contains the cut point $c$?
Why does the notation $(a,b]$ not imply that the set is open?

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