The most important point in this section is to understand the definitions of open and closed sets, and to develop a good intuitive feel for what these sets are like. This requires some understanding of the notions of boundary, interior, and closure.
In fact, we will see soon that many sets can be recognized as open or closed, by the nature of their description, without appealing to \((\varepsilon, \delta)\) arguments. Just like limits of functions in single variable calculus, we want to ensure that they have a solid definition, and that we can explain with \((\varepsilon, \delta)\) if we need to. But we won’t always need to.
Some proofs are given here and in the lectures. These sorts of proofs are good practice for theorem-proving skills, and straightforward proofs of this sort would be reasonable test questions.
Basic terminology and notation
The interval is a fundamental object in \(\R\). Most of the theorems in one variable involve functions with an open or closed interval as a domain, so we would like to generalize the concept of an interval. The first way to do it is to think of the interval as the set of points close to the centre of the interval. In \(\R^n\), we call this an open ball.
For \(\mathbf a\in \R^n\) and \(r>0\), the open ball with centre \(\mathbf a\) and radius \(r\) is the set \[\{ \mathbf x \in \R^n : |\mathbf x - \mathbf a|< r\}.\] It is denoted \(B(\mathbf a;r)\).
We may say “ball” instead of “open ball”, and may call this the neighbourhood of radius \(r\) around \(\mathbf a\).
For \(\mathbf a\in \R^n\) and \(r>0\), the sphere with centre \(\mathbf a\) and radius \(r\) is the set of points that are exactly distance \(r\) from \(\mathbf a\): \[
\{ \mathbf x \in \R^n : |\mathbf x - \mathbf a| = r\}.
\]
Although “sphere” and “ball” may be used interchangeably in ordinary English, in mathematics they have different meanings. We will not use sphere often enough to warrant a special notation. The union of an open ball and its sphere is called the closed ball with centre \(\mathbf a\) and radius \(r\), \(\overline B(\mathbf a;r)=\{ \mathbf x \in \R^n : |\mathbf x - \mathbf a|\leq r\}\), named like a closed interval.
We will write \(\mathbf 0\), in boldface, to denote the origin in \(\R^n\).
A set \(S\subseteq \R^n\) is bounded if there exists some \(r>0\) such that \(S\subseteq B({\mathbf 0};r)\). A set is unbounded if and only if it is not bounded.
Compare this to your definition of bounded sets in \(\R\).
Interior, boundary, and closure
Assume that \(S\subseteq \R^n\) and that \(\mathbf x\) is a point in \(\R^n\). Imagine you zoom in on \(\mathbf x\) and its surroundings with a microscope that has unlimited powers of magnification. This is an experiment that is beyond the reach of current technology but can be carried out with perfect accuracy in your mind. One of three possibilities must occur:
There is some magnification beyond which you see only points that belong to \(S\). More precisely, \[\begin{equation}\label{interior}\exists \varepsilon>0 : B(\mathbf x;\varepsilon)\subseteq S. \end{equation}\]
In this case, we say that \(\mathbf x\) belongs to the interior of \(S\). We write \(S^{int}\) for the set of all such points.
There is some magnification beyond which you see only points that do not belong to \(S\) (or equivalently, that belong to \(S^c\)). More precisely, \[\begin{equation}\label{compint}\exists \varepsilon>0 : B(\mathbf x;\varepsilon)\subseteq S^c.\end{equation}\]
None of the above: no matter how much you turn up the magnification, you always see both some points that belong to \(S\), and some that do not. More precisely, \[\begin{equation}\label{boundary}
\forall \varepsilon>0 : \left(B(\mathbf x;\varepsilon)\cap S\ne \emptyset \ \text{ and } \ B(\mathbf x;\varepsilon)\cap S^c\ne \emptyset\ \right).\end{equation}\]
In this case, we say that \(\mathbf x\) belongs to the boundary of \(S\). We write \(\partial S\) for the set of all such points.
We say that \(\mathbf x\) belongs to the closure of \(S\), and we write \(\mathbf x \in \overline S\), if \(\mathbf x\) is in the interior or the boundary, that is, \(\overline S = S^{int}\cup\partial S.\) This is the same as saying that \[\overline S = \{ \mathbf x \in \R^n : \forall \varepsilon>0, \quad B(\mathbf x;\varepsilon,)\cap S\ne \emptyset\}.\]
What about Case 2 above? Recall that if \(S\subseteq \R^n\), then the complement of \(S\), denoted \(S^c\), is \(\{ \mathbf x\in \R^n : \mathbf x \not\in S\}.\) In Case 2, there is an open ball around \(\mathbf x\) that is contained entirely outside of \(S\), so \(\mathbf x\) is in the interior of the complement of \(S\), in symbols, \(\mathbf x \in (S^c)^{int}\). Equivalently, \(\mathbf x \in \left(\overline S\right)^c\). Try to show that these are equivalent.
Here are some basic properties of interiors, boundaries and closures. Try to write out formal proofs of these before looking at the proofs.
For any \(S\subseteq \R^n\), \[
S^{int} \subseteq S \subseteq \overline S.
\] In particular, every point of \(S\) is either an interior point or a boundary point.
Details
To prove that \(S^{int}\subseteq S\), consider an arbitrary point \(\mathbf x \in S^{int}\). By definition of interior, there exists \(\varepsilon>0\) such that \(B(\mathbf x;\varepsilon)\subseteq S\). Since \(\mathbf x\in B(\mathbf x;\varepsilon)\), it follows that \(\mathbf x\in S\). Since \(\mathbf x\) was an arbitrary point of \(S^{int}\), it follows that \(S^{int}\subseteq S\).
Next, consider an arbitrary point \(\mathbf x\) of \(S\). Then for every \(\varepsilon>0\), both \(\mathbf x \in B(\mathbf x;\varepsilon)\) and \(\mathbf x \in S\) are true. Hence \(B(\mathbf x;\varepsilon)\cap S\ne \emptyset\) for every \(\varepsilon>0\). This says that \(\mathbf x\in \overline S\). Since \(\mathbf x\) was an arbitrary point of \(S\), it follows that \(S\subseteq \overline S\).
Finally, the statement that \(S\subseteq \overline S\) says exactly that every point of \(S\) is either an interior point or a boundary point, since \(\overline S = S^{int}\cup \partial S\).
For any \(S\subseteq \R^n\), \[
\partial S = \partial (S^c).
\]
Details
First we claim that \[\begin{equation}\label{cc}
(S^c)^c = S.
\end{equation}\]
This is probably familiar from earlier classes, and can be checked by unwinding the definitions: \[\begin{align*}
\mathbf x\in (S^c)^c &\quad\iff\qquad \mathbf x\not\in S^c = \{ \mathbf y\in \R^n : \mathbf y\not\in S\} \\\
&\quad\iff\qquad\mathbf x\in S
\end{align*}\] This proves \(\eqref{cc}\). Next, we use \(\eqref{cc}\) to deduce that \[\begin{align*}
\mathbf x \in \partial(S^c)
&\iff \
\forall \varepsilon>0, \ \ B(\mathbf x; \varepsilon)\cap S^c\ne \emptyset \ \text{ and } \ B(\mathbf x; \varepsilon)\cap (S^c)^c\ne \emptyset\ \\\
&\iff \
\forall \varepsilon>0, \ \ B(\mathbf x; \varepsilon)\cap S^c\ne \emptyset \ \text{ and } \ B(\mathbf x; \varepsilon)\cap S\ne \emptyset\ \\\
&\iff \mathbf x\in \partial S
\end{align*}\] This completes the proof.
Examples
What are the interior, boundary, and closure of an open ball \(B(\mathbf a, r)\), for \(\mathbf a\in \R^n\) and \(r>0\)?
Solution: Let \(S = B(\mathbf a; r)\). By applying the definitions, we can see that \(S^{int} = S\), and \(\partial S = \{\mathbf x\in \R^n : |\mathbf x - \mathbf a|=r\}\). Thus \[\overline S = S^{int}\cup \partial S = \{\mathbf x\in \R^n : |\mathbf x - \mathbf a| \le r\},\] the closed ball.
If we want to prove these (not recommended, for the assertion about \(\partial S\)), we can do so as follows:
\(S^{int}= S\)
We already know from Theorem 1 that \(S^{int}\subseteq S\), so we only have to prove that \(S\subseteq S^{int}\). To do this, we must prove that \(\forall \mathbf x\in S\), condition \(\eqref{interior}\) holds. So, pick \(\mathbf x\in S\). Let \(s = |\mathbf x-\mathbf a|\). By definition of \(S\), we know that $ s < r $. We claim (motivated by drawing a picture) that if we let \(\varepsilon = r-s\), then \(B(\mathbf x; \varepsilon)\subseteq S\). To prove it, consider any \(\mathbf y \in B(\mathbf x; \varepsilon)\). By the triangle inequality, \[|\mathbf y-\mathbf a| = |(\mathbf y - \mathbf x) + (\mathbf x-\mathbf a)|\le |\mathbf y-\mathbf x| +|\mathbf x-\mathbf a|< \varepsilon+s\] since \(|\mathbf x-\mathbf a| = s\) and \(| \mathbf y - \mathbf x | < \varepsilon\) for \(\mathbf y \in B(\mathbf x; \varepsilon)\). Since we chose \(\varepsilon = r-s\), it follows that \(\mathbf y\in B(\mathbf a;r) = S\). Since \(\mathbf y\in B(\mathbf x;\varepsilon)\) was arbitrary, it follows that \(B(\mathbf x; \varepsilon)\subseteq S\), and hence that \(\mathbf x \in S^{int}\). Since \(\mathbf x\) was an arbitrary point of \(S\), this shows that \(S\subseteq S^{int}\).
\(\partial S\) is the sphere
Let \(T=\{ \mathbf x \in \R^n : |\mathbf x - \mathbf a| = r\}.\) This proof is pretty complicated, because there are a lot of details to keep straight. We must prove that \(\partial S \subseteq T\) and that \(T\subseteq \partial S\). * \(\partial S\subseteq T\): We already know that if \(|\mathbf x-\mathbf a|<r\), then \(\mathbf x\in S^{int}\), and thus \(\mathbf x\not\in \partial S\).
* Essentially the same argument shows that if \(|\mathbf x-\mathbf a|>r\), then \(\mathbf x\in (S^c)^{int}\), and thus \(\mathbf x\not\in \partial S\). This completes the proof that \(\partial S\subseteq T\).
To show \(T\subseteq \partial S\), let \(\mathbf x\in T\). We must check that that for every \(\varepsilon>0\), \(B(\mathbf x, \varepsilon)\) intersects both \(S\) and \(S^c\). Thus we consider: * \(B(\mathbf x, \varepsilon)\cap S^c\ne \emptyset\). This is clear, since \(\mathbf x\in T \subseteq S^c\). * \(B(\mathbf x, \varepsilon)\cap S\ne \emptyset\). This is the hardest point. One way to do it is to specify a point \(\mathbf y\) that belongs to both \(S\) and \(B(\mathbf x; \varepsilon)\). This can be done by choosing \(\mathbf y\) on the line segment from \(\mathbf a\) to \(\mathbf x\), that is \(\mathbf y = \mathbf a + t(\mathbf x - \mathbf a)\) for \(0<t<1\), and then taking \(t>1-\varepsilon\).
Should you practice rigorously proving that the interior/boundary/closure of a set is what you think it is? This will mostly be unnecessary, due to an easy test that we will introduce in Section 1.2.3. In general, we do not recommend spending a lot of time proving these the long way.
On the other hand, the proof that every point of an open ball is an interior point is fundamental, and you should understand it well.
For each of the sets below, determine (without proof) the interior, boundary, and closure. Some of these examples, or similar ones, may be discussed in the lectures.
Hint for 5,6,7It is useful to keep in mind that every open interval \((a,b)\subseteq \R\) contains both rational and irrational numbers.
\(\quad S = \left\{(x,y)\in \R^2 : x>0 \text{ and } y\ge 0\right\}\).
\(\quad S = \left\{ \left(\dfrac 1n, \dfrac 1{n^2}\right) \in \R^2 : n\text{ is a positive integer}\right\}.\)
\(\quad S = \left\{ (x,y)\in \R^2 : y = x^2 \right\}\).
\(\quad S = \left\{ (x,y,z)\in \R^3 : z > x^2 + y^2 \right\}\).
\(\quad S = \left\{ x\in (0,1) : x\text{ is rational} \right\}\).
\(\quad S = \left\{ (x,y)\in \R^2 : x\text{ is rational} \right\}\).
\(\quad S = \left\{ \mathbf x \in \R^3 : 0< |\mathbf x| < 1, \ |\mathbf x| \text{ is irrational} \right\}\).
Open and closed sets
Considering only open or closed balls will not be general enough for our domains. To generalize open and closed intervals, we will consider their boundaries and interiors. By our definition, the boundary of an interval is the set of two endpoints. Then we categorize types of intervals by whether they contain all of their boundary points or not. The closed interval \([a,b]\) contains all of its boundary points, while the open interval \((a,b)\) contains none of them. We generalize these terms to sets in \(\R^n\):
A set \(S\) is open if \(S = S^{int}\).
A set \(S\) is closed if \(S = \overline S\).
In Section 1.2.3, we will see how to quickly recognize many sets as open or closed.
Contrary to what the names “open” and “closed” might suggest,
some sets are both open and closed (try to find them), and
many sets are neither open nor closed, if they contain some boundary points and not others.
In this class, we will mostly see open and closed sets. For example,
when we study differentiability in Section 2.1, we will frequently consider either
differentiable functions whose domain is an open set, or
any function whose domain is a closed set, but that is differentiable at every point in the interior.
when we study optimization problems in Section 2.8, we will normally find it useful to assume that the domain is closed.
Here are alternate characterizations of open and closed sets that are often useful in proofs.
\[\begin{align*}
S \text{ is open}
&\quad \iff \quad \text{ every point of $S$ is an interior point}
\\\
&\quad \iff \quad \forall \mathbf x\in S \ \ \exists \varepsilon >0\text{ such that }B(\mathbf x; \varepsilon)\subseteq S
\quad\end{align*}\]
Details
First, if \(S\) is open, then \(S = S^{int}\), which certainly implies that \(S\subseteq S^{int}\), or in other words that every point of \(S\) is an interior point.
Conversely, assume that every point of \(S\) is an interior point, or in other words that \(S\subseteq S^{int}\). We know from Theorem 1 above that \(S^{int}\subseteq S\). Combining these, we conclude that \(S=S^{int}\). This completes the proof of the first “\(\iff\)” in the statement of the theorem. The second “\(\iff\)” follows directly from the definition of interior point.
\[\begin{align*}
S \text{ is closed}
&\quad \iff \quad \text{ every boundary point of $S$ belongs to $S$}
\\\
&\quad \iff \quad S^c \text{ is open}.
\end{align*}\]
Details
By definition, if \(S\) is closed, then \(S = \overline S = S^{int}\cup \partial S\). This certainly implies that \(\partial S\subseteq S\), or in other words that every boundary point of \(S\) belongs to \(S\).
Conversely, assume that \(\partial S\subseteq S\). We know from Theorem 1 above that \(S^{int}\subseteq S\). Combining these, we conclude that \(\overline S\subseteq S\). Again using Theorem 1, we recall that \(S\subseteq \overline S\). It follows that \(\overline S = S\), and hence that \(S\) is closed. This completes the proof of the first “\(\iff\)” in the statement of the theorem.
Next, since \(\partial S = \partial S^c\) and every point of \(S^c\) belongs either to \((S^c)^{int}\) or \(\partial(S^c)\), \[\begin{align*}
S\text{ is closed } &\iff \partial S \subseteq S \iff \partial (S^c) \subseteq S \\\
&\iff
\text{ no point of $S^c$ is a boundary point } \iff S^c\text{ is open}.
\end{align*}\] This completes the proof.
More examples
Every open ball \(B(\mathbf a; r)\) is an open set. Although this sounds obvious, the proof given above requires a few steps.
The above definitions (open ball, open set, closed set …) all make sense when \(n=1\), that is, for subsets of \(\R\). In this case,
an open interval \((a,b)\) is an open set. This is also true for intervals of the form \((a,\infty)\) or \((-\infty, b)\).
A closed interval \([a,b]\) is a closed set.
Try to find other examples of open sets and closed sets in \(\R\).
Give an example of a set \(S\subseteq \R^n\) that is both open and closed. Can you think of two different sets with this property? How about three?
HintIf \(S\) is open then \(\partial S \cap S = \emptyset\). This is a consequence of Theorem 2. On the other hand, if \(S\) is closed, then \(\partial S \subseteq S\). How can these both be true at once?
What is an example of a set \(S\subseteq \R^n\) that is neither open nor closed?
One example\(B(\mathbf 0; 1)\cup \partial B(\mathbf 0; 2)\) contains some of its boundary but not all of it, so it isn’t open or closed.
Problems
For the sets in problems 1-9, determine:
the interior, boundary, and closure,
whether they are bounded or unbounded, and
whether they are open, closed, neither, or both.
No proof is necessary.
\(\quad S = \{ \mathbf x \in \R^n : |\mathbf x|<1\}\).
\(\quad S = \{(x,y)\in \R^2 : x>0 \text{ and } y\ge 0\}\).
\(\quad S = \left\{ \left(\dfrac 1n, \dfrac 1{n^2}\right) \in \R^2 : n\text{ is a positive integer}\right\}.\)
\(\quad S = \{ (x,y)\in \R^2 : y = x^2 \}\).
\(\quad S = \{ (x,y,z)\in \R^3 : z > x^2 + y^2 \}\).
\(\quad S := \{ x\in\R : x\in (0,1)\text{ and } x\text{ is rational} \}\).
\(\quad S = \{ (x,y)\in \R^2 : x\text{ is rational } \}\).
\(\quad S = \{ \mathbf x \in \R^3 : 0< |\mathbf x| < 1, \ |\mathbf x| \text{ is irrational} \}\).
\(\quad S = \{ \mathbf x \in \R^n : |\mathbf x| = 2^{-j} \text{ for some }j\in {\mathbb N}\}\).
Can a set be both open and closed at the same time?
Must a set be either open or closed?
Can a set be both bounded and unbounded at the same time?
Must a set be either bounded or unbounded?
Advanced
Prove that if \(A, B\) are open subsets of \(\R^n\) then \(A\cup B\) and \(A\cap B\) are open.
Deduce from this and de Morgan’s laws that if \(A, B\) are closed subsets of \(\R^n\) then \(A\cup B\) and \(A\cap B\) are closed.
Hint De Morgan’s laws state that \((A\cup B)^c = A^c \cap B^c\) and \((A\cap B)^c = A^c \cup B^c\).
If \(A_1, A_2, \ldots\) is a sequence of subsets of \(\R^n\), then we define \[
\bigcup_{j\ge 1} A_j := \{ \mathbf x\in \R^n : \exists j \ge 1\text { such that }\mathbf x\in A_j \}.
\] Prove that if \(A_j\) is open for every \(j\), then so is \(\bigcup_{j\geq 1} A_j\).
If \(A_1, A_2, \ldots\) is a sequence of subsets of \(\R^n\), then we define \[
\bigcap_{j\ge 1} A_j := \{ \mathbf x\in \R^n : \mathbf x\in A_j \,\forall j\ge 1 \}.
\]
Show that \[\bigcap_{j\ge 1} B({\mathbf 0}; 1+ 2^{-j}) = \overline{B}({\mathbf 0}; 1).\]
Is it true that if \(A_j\) is open for every \(j\), then \(\bigcap_{j\ge 1} A_j\) must be open?
Is it true that if \(A_j\) is closed for every \(j\), then \(\bigcup_{j\ge 1} A_j\) must be closed?
Hint It may be relevant to note that \(\left(\bigcup_{j\ge 1} A_j\right)^c = \bigcap_{j\ge 1} A_j^c\).
Assume that \(A\) is a nonempty open subset of \(\R\), and let \[
S = \{ (x,0) : x\in A \} \subseteq \R^2.
\]That is, \(S\) is a copy of \(A\) on the \(x\)-axis in \(\R^2\). Is \(S\) open, closed, or neither? Prove that your answer is correct.
Assume that \(A_1\) and \(A_2\) are nonempty open subsets of \(\R\), and let \[
S := \{ (x,y) : x\in A_1, y\in A_2 \} \subseteq \R^2.
\] Is \(S\) open, closed, or neither? Prove that your answer is correct.
