Finite Intersections of Open Sets

Combining Finitely Many Open Conditions

An open set provides room around each of its points: every point in the set has an open interval around it that remains inside the set. When a point must satisfy several open conditions at once, each condition supplies its own interval. With only finitely many conditions, the point has room to satisfy them all simultaneously. The radius of that shared interval may be smaller than every radius considered separately, but it can still be chosen positive.

The Finite Intersections of Open Sets Theorem was established earlier in this course. We will use that result and look more closely at what finiteness contributes. In particular, we will calculate intersections of intervals and show that the interior operation commutes with finite intersections of arbitrary sets.

Finite Intersections and a Shared Radius

For sets $U_1,\ldots,U_m$, their intersection contains exactly the points that belong to every one of the sets:

\bigcap_{j=1}^{m}U_j = \{x\in\mathbb{R}:x\in U_j\text{ for every }j\in\{1,\ldots,m\}\}.

Thus, proving that $x$ belongs to an intersection requires checking every set in the family. By contrast, to prove that $x$ does not belong, it is enough to find one set that excludes it.

Theorem (Finite Intersections of Open Sets, recalled): If $U_1,\ldots,U_m$ are open subsets of $\mathbb{R}$, where $m$ is a positive integer, then $\bigcap_{j=1}^{m}U_j$ is open. The intersection of an empty family is taken to be $\mathbb{R}$, which is also open.

The local radius calculation explains how to use this theorem. Suppose $x$ belongs to every $U_j$, and each $U_j$ is open. For each index $j$, choose $r_j>0$ such that $B_{r_j}(x)\subseteq U_j$. Since there are only finitely many positive radii, their minimum is positive:

r=\min\{r_1,\ldots,r_m\}>0.

Because $r\leq r_j$ for every $j$, we have $B_r(x)\subseteq B_{r_j}(x)\subseteq U_j$ for every $j$. Therefore $B_r(x)\subseteq\bigcap_{j=1}^{m}U_j$. The crucial point is that the radius may depend on $x$, but at a given point a finite list of radii has a positive minimum. There is no need for one radius to work at every point of the intersection.

Worked Example: Finding the Common Neighborhood Radius

Let $U_1=(-2,5)$, $U_2=(-1,3)$, and $U_3=(0,4)$. Consider the point $x=1$. The distances from $1$ to the endpoints of each interval show that the following radii work:

B_2(1)\subseteq U_1,\qquad B_1(1)\subseteq U_2,\qquad B_1(1)\subseteq U_3.

Indeed, $B_2(1)=(-1,3)$, which is contained in $(-2,5)$. Also, $B_1(1)=(0,2)$, which is contained in both $(-1,3)$ and $(0,4)$. The minimum of the three radii is $1$, so

B_1(1)=(0,2)\subseteq U_1\cap U_2\cap U_3.

This radius is valid, but it is not the largest possible one. The intersection is $(0,3)$, so a ball centered at $1$ can have radius at most $1$ while remaining inside it. Here the minimum-radius calculation happens to give that largest radius. In general, the radii chosen for the individual sets need not be optimal; their minimum is a convenient guaranteed choice.

Calculating Intersections of Open Intervals

For a finite family of bounded open intervals, the intersection can be calculated by comparing their endpoints. To remain in every interval, a point must be greater than every left endpoint and less than every right endpoint. The strongest lower restriction is the largest left endpoint, and the strongest upper restriction is the smallest right endpoint.

Theorem (Finite Intersection of Bounded Open Intervals): Let $I_j=(a_j,b_j)$ for $j=1,\ldots,m$, where $a_j<b_j$. Put $a=\max\{a_1,\ldots,a_m\}$ and $b=\min\{b_1,\ldots,b_m\}$. If $a<b$, then $\bigcap_{j=1}^{m}I_j=(a,b)$. If $a\geq b$, then the intersection is empty.

Proof. A real number $x$ belongs to every $I_j$ exactly when $a_j<x<b_j$ for every $j$. The inequalities $a_j<x$ for every $j$ are equivalent to $a<x$, since $a$ is the largest left endpoint. Likewise, $x<b_j$ for every $j$ is equivalent to $x<b$, since $b$ is the smallest right endpoint. Consequently,

x\in\bigcap_{j=1}^{m}I_j \quad\Longleftrightarrow\quad a<x<b.

If $a<b$, these are exactly the conditions for $x\in(a,b)$. If $a\geq b$, no real $x$ can satisfy both $a<x$ and $x<b$, so the intersection is empty. This proves both cases. $\square$

Worked Example: The Most Restrictive Endpoints

Consider the three intervals

I_1=(-3,4),\qquad I_2=(1,6),\qquad I_3=(-1,5).

Their left endpoints are $-3$, $1$, and $-1$, so the largest is $1$. Their right endpoints are $4$, $6$, and $5$, so the smallest is $4$. Since $1<4$, the theorem gives

I_1\cap I_2\cap I_3=(1,4).

To check the conclusion directly, a point in $(1,4)$ is greater than $-3$, $1$, and $-1$, and is less than $4$, $6$, and $5$. It therefore belongs to all three intervals. Conversely, a point in all three must be greater than $1$, because it belongs to $I_2$, and less than $4$, because it belongs to $I_1$. Thus it must lie in $(1,4)$, verifying the equality in both directions.

Worked Example: When the Endpoint Restrictions Do Not Overlap

Let $J_1=(-4,-1)$ and $J_2=(-2,3)$. The largest left endpoint is $-2$, while the smallest right endpoint is $-1$, so their intersection is $(-2,-1)$. If instead $J_2$ is replaced by $K_2=(0,3)$, the largest left endpoint is $0$ and the smallest right endpoint is $-1$. Since $0\geq-1$, the theorem gives

(-4,-1)\cap(0,3)=\varnothing.

Indeed, a point in the first interval must be less than $-1$, while a point in the second must be greater than $0$; no real number can meet both requirements. The equality case in the theorem matters as well: if the intervals were $(-4,0)$ and $(0,3)$, the largest left endpoint and smallest right endpoint would both be $0$. The only possible shared endpoint is excluded from both open intervals, so their intersection is empty.

Interiors and Finite Intersections

The same finite-intersection principle has a useful consequence for sets that are not themselves open. Recall that $\operatorname{int}(E)$ is the largest open subset of $E$. A point belongs to the interior of an intersection precisely when it can be included in an open set that stays inside every one of the sets being intersected. For finitely many sets, this gives an exact identity.

Theorem (Interior of a Finite Intersection): For subsets $E_1,\ldots,E_m$ of $\mathbb{R}$, where $m$ is a positive integer, $$ \operatorname{int}\left(\bigcap_{j=1}^{m}E_j\right) = \bigcap_{j=1}^{m}\operatorname{int}(E_j). $$

Proof. Write $E=\bigcap_{j=1}^{m}E_j$. Since $E\subseteq E_j$ for each $j$, monotonicity of the interior gives $\operatorname{int}(E)\subseteq\operatorname{int}(E_j)$ for every $j$. Therefore

\operatorname{int}(E)\subseteq\bigcap_{j=1}^{m}\operatorname{int}(E_j).

For the reverse inclusion, each $\operatorname{int}(E_j)$ is open. The Finite Intersections of Open Sets Theorem implies that $G=\bigcap_{j=1}^{m}\operatorname{int}(E_j)$ is open. Also, $\operatorname{int}(E_j)\subseteq E_j$ for each $j$, so $G\subseteq\bigcap_{j=1}^{m}E_j=E$. Thus $G$ is an open subset of $E$. By the largest-open-subset property of the interior, $G\subseteq\operatorname{int}(E)$. Combining the two inclusions proves the identity. $\square$

Worked Example: Taking the Interior After Intersecting

Let $E=[-2,1]\cup\{4\}$ and $F=[0,3]\cup\{4\}$. Their interiors are

\operatorname{int}(E)=(-2,1),\qquad \operatorname{int}(F)=(0,3).

Their intersection is $E\cap F=[0,1]\cup\{4\}$. The point $4$ is isolated from the interval piece and cannot have an open interval around it contained in $E\cap F$. The endpoints $0$ and $1$ also cannot belong to the interior: every interval around $0$ contains points less than $0$, and every interval around $1$ contains points greater than $1$. Points strictly between $0$ and $1$ do have open intervals contained in the intersection. Hence

\operatorname{int}(E\cap F)=(0,1) = \operatorname{int}(E)\cap\operatorname{int}(F).

The theorem explains why this calculation is not a coincidence. For any finite collection of sets, taking their interiors first and then intersecting gives the same result as intersecting first and then taking the interior.

Why Finiteness Matters

In the common-radius calculation, finitely many positive radii have a positive minimum. An infinite collection of positive radii need not have a positive lower bound. That difference is one reason arbitrary intersections of open sets need not be open.

Worked Example: Infinitely Many Open Intervals with a Nonopen Intersection

For each positive integer $n$, let $U_n=(-1/n,1/n)$. Each $U_n$ is open, and $0$ belongs to every $U_n$. We claim that

\bigcap_{n=1}^{\infty}U_n=\{0\}.

If $x=0$, then $-1/n<0<1/n$ for every positive integer $n$, so $0$ belongs to the intersection. If $x\neq0$, choose a positive integer $n>1/|x|$. Then $1/n<|x|$, so $x\notin(-1/n,1/n)$. Thus no nonzero $x$ lies in every interval, proving the claimed equality.

The singleton $\{0\}$ is not open: every open interval around $0$ contains nonzero real numbers. There is also no positive radius that works simultaneously for all the $U_n$. Given any $r>0$, choose $n>1/r$. Then $1/n<r$, so the ball $B_r(0)=(-r,r)$ is not contained in $U_n=(-1/n,1/n)$. The finite-radius argument cannot be extended by simply taking a minimum over infinitely many radii.

The finite-intersection theorem guarantees openness when the number of open sets is finite; it does not make a claim about arbitrary intersections. Similarly, the identity for interiors proved above relies on finiteness in its use of the finite-intersection theorem for open sets. When working with intersections, first check how many sets are involved, then identify the conditions that every point must satisfy. For intervals, compare the most restrictive endpoints; for neighborhoods, take the minimum of the finitely many radii; and for interiors, use the finite-intersection identity only when its hypotheses apply.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

In a finite intersection, what must be true of a point for it to belong to the intersection?
If a point lies in finitely many open sets, why can the individual neighborhood radii be replaced by one positive radius?
Calculate $(-5,2)\cap(-1,6)\cap(0,4)$ by comparing the largest left endpoint and the smallest right endpoint.
Why is the intersection of $(-3,1)$ and $(1,5)$ empty, even though the intervals meet at an endpoint?
State the identity relating the interior of a finite intersection to the intersection of the interiors.
Why does the family $(-1/n,1/n)$, for positive integers $n$, not contradict the theorem on finite intersections of open sets?