Open Versus Closed

Two Different Ways a Set Can Behave

The previous tutorial used examples and the Sequential Characterization of Closed Sets to test whether a set contains the limits of its convergent sequences. Openness asks a different question: does every point of the set have a whole neighborhood that stays inside the set? These are distinct conditions. A set can be open without being closed, closed without being open, both, or neither.

Recall that \(U\subseteq\mathbb{R}\) is open if, for each \(x\in U\), there is an \(r>0\) such that \(B_r(x)\subseteq U\). A set \(F\subseteq\mathbb{R}\) is closed if it contains the limit of every convergent sequence of its points. The Sequential Characterization of Closed Sets says this is equivalent to closedness in the usual sense.

This distinction is easy to lose when looking only at familiar intervals. Open intervals are open but not closed, while closed intervals are closed but not open when they have distinct endpoints. The endpoint behavior explains these examples, but the distinction applies to arbitrary subsets of the real line as well.

Comparing the Four Possibilities

The empty set and the whole real line each satisfy both conditions. The empty set has no points that could violate the definition of openness, and every sequence of its points is vacuously covered by the closedness condition. Every ball around a real number is contained in \(\mathbb{R}\), and every real limit belongs to \(\mathbb{R}\). The other two possibilities are illustrated by intervals and will be proved in the examples below.

Open Sets Have No Isolated Points

A point \(x\in E\) is called an isolated point of \(E\) if some ball around \(x\) contains no other point of \(E\); in other words, there is an \(r>0\) such that \(B_r(x)\cap E=\{x\}\). The definition of an open set rules out this behavior on the real line.

Proof. Let \(x\in U\). Since \(U\) is open, there is an \(r>0\) such that \(B_r(x)\subseteq U\). Let \(\varepsilon>0\) be arbitrary, and choose \(t=\min\{r,\varepsilon\}/2\). Then \(t>0\), \(t<r\), and \(t<\varepsilon\). The point \(y=x+t\) is distinct from \(x\), and \(|y-x|=t<r\), so \(y\in B_r(x)\subseteq U\). Also, \(|y-x|=t<\varepsilon\). Thus every ball around \(x\), however small, contains a point of \(U\) other than \(x\). Therefore \(x\) is not isolated in \(U\). Since \(x\) was arbitrary, \(U\) has no isolated points. \(\square\)

This theorem is specific to the real line, where there are always other real numbers arbitrarily near \(x\). It does not say that a closed set has no isolated points. For example, a singleton is closed, and its one point is isolated. The property of being open and the property of being closed should therefore not be understood as opposite ends of a single scale.

The Only Sets That Are Both Open and Closed

The examples above include sets that are both open and closed: \(\varnothing\) and \(\mathbb{R}\). In fact, there are no others. The proof uses the order structure of the real line. If a nonempty proper set and its outside both had the neighborhood protection given by openness, a supremum would identify a point where that protection fails.

Proof. Suppose, to the contrary, that \(A\) is both open and closed, nonempty, and not equal to \(\mathbb{R}\). Choose \(p\in A\) and \(q\notin A\). If \(p>q\), replace \(A\) by \(-A=\{-x:x\in A\}\), and replace \(p,q\) by \(-p,-q\). Reflection preserves openness because it takes balls to balls of the same radius: \(B_r(-x)=\{-y:y\in B_r(x)\}\). It preserves closedness as well, since the negatives of a convergent sequence converge to the negative of its limit. We may therefore assume \(p<q\).

The set \(S=A\cap[p,q]\) is nonempty because \(p\in S\), and it is bounded above by \(q\). Let \(c=\sup S\). For every positive integer \(n\), the number \(c-1/n\) cannot be an upper bound for \(S\), since it is strictly less than the least upper bound \(c\). Choose \(s_n\in S\) with \(c-1/n<s_n\leq c\). It follows that \(0\leq c-s_n<1/n\), so \(s_n\to c\). Each \(s_n\) belongs to \(A\), and \(A\) is closed. The Sequential Characterization of Closed Sets gives \(c\in A\).

If \(c=q\), then \(q\in A\), contradicting the choice of \(q\). If \(c<q\), openness of \(A\) at \(c\) gives an \(r>0\) with \(B_r(c)\subseteq A\). Choose \(t=\min\{r,q-c\}/2\), which is positive and less than both \(r\) and \(q-c\). Then \(c+t\in B_r(c)\subseteq A\), and \(p\leq c<c+t<q\), so \(c+t\in S\). This contradicts that \(c\) is an upper bound for \(S\). Both possibilities lead to contradictions. Hence no nonempty proper subset of \(\mathbb{R}\) can be both open and closed. Since \(\varnothing\) and \(\mathbb{R}\) do satisfy both definitions, the theorem follows. \(\square\)

How to Choose the Right Test

When deciding whether a particular set is open or closed, use the condition that matches the question. To disprove openness, it is enough to find one point of the set for which every ball contains a point outside it. To disprove closedness, it is enough to find a convergent sequence of points in the set whose limit lies outside it. Conversely, proving either property requires checking its condition for all relevant points or sequences.

A frequent pitfall is to infer closedness from the word “closed interval,” or openness from the word “open interval,” without checking what happens at endpoints. Another is to suppose a set must be one or the other. The interval \([0,2)\) is a direct counterexample: it is neither. Finally, the theorem about sets that are both open and closed is not an assertion that openness and closedness are opposites. The empty set and the whole line have both properties, and the two properties have different definitions and different tests.