Open sets and closed sets in \(\mathbb{R}^n\)

Open sets

In \(\mathbb{R}\), we have the notion of an open interval such as \((a, b) = \{x \in \mathbb{R} | a < x < b\}\). We want to extend this idea to apply to \(\mathbb{R}^n\). We also introduce the notions of bounded sets and closed sets in \(\mathbb{R}^n\).

Recall that the \(\varepsilon\)-neighbor of a point \(x\in\mathbb{R}^n\) is defined as \(N_{\varepsilon}(x) = \{y \in \mathbb{R}^n | d(x, y) < \varepsilon \}\) where \(d(x,y)\) is the distance between \(x\) and \(y\).

Definition (Open set)

A subset \(U\) of \(\mathbb{R}^n\) is said to be an open set if the following holds:

\[\forall x \in U ~ \exists \delta > 0 ~ (N_{\delta}(x) \subset U).\tag{Eq:OpenSet}\]

That is, for every point in an open set \(U\), we can always find an open ball centered at that point, that is included in \(U\). See the following figure.

Perhaps, it is instructive to see what is not an open set. Negating (Eq:OpenSet), we have \[\exists x \in U ~ \forall \delta > 0 ~ (N_{\delta}(x) \not\subset U).\] This means that there is some point in \(U\) such that its neighbor cannot be included in \(U\) (i.e., some part of the neighbor crosses the "boundary" to go out of \(U\).) 

Example. The open interval \((0, 1)\) in \(\mathbb{R}\) is an open set. To see this, take any point \(x \in (0, 1)\) and let \(\delta = \min\{x, 1 - x\}\) (note: \(x>0\) and \(1-x>0\)). Then, we have the \(\delta\)-neighbor \(N_{\delta}(x) = (x - \delta, x + \delta) \subset (0, 1)\). □

Example. The closed interval \([0, 1]\) is not an open set. To see this, let \(x = 1 \in [0, 1]\). For any \(\delta > 0\), we have the neighbor \(N_{\delta}(1) = (1 - \delta, 1 + \delta) \not\subset [0, 1]\) because \(1 + \delta > 1\).  □

Definition (Path-connected)

1. For each \(i = 1, 2, \cdots, n\), let \(\sigma_i(t)\) be a continuous function on \([0, 1]\). Then the \(n\)-tuple of these functions, \(\sigma(t) = (\sigma_1(t), \sigma_2(t), \cdots, \sigma_n(t))\) is a point in \(\mathbb{R}^n\), parameterized by \(t\in [0,1]\). The set defined by \[\sigma = \{\sigma(t) \mid t \in [0, 1]\}\] is called a path in \(\mathbb{R}^n\). The points \(\sigma(0)\) and \(\sigma(1)\) are called the initial and terminal points, respectively, and both are collectively called the endpoints.
2. If the path \(\sigma\) in \(\mathbb{R}^n\) is included in the subset \(S\) of \(\mathbb{R}^n\), that is, \(\sigma \subset S\), then \(\sigma\) is said to be a path in \(S\).
3.  For the subset \(S\) of \(\mathbb{R}^n\), for any \(x, y\in S\), if there exists at least one path \(\sigma\) with \(x\) and \(y\) being its endpoints, \(S\) is said to be path-connected.

Example. Let \(o = (0, 0)\), \(a = (3, 0)\) in \(\mathbb{R}^2\) and consider the union \(S\) of  1-neighbors (i.e., \(\varepsilon =1\)): \(S = N_{1}(o)\cup N_{1}(a)\) (Draw pictures!). No matter how we connect the two points \(o\) and \(a\) by a path (continuous line or curve), there are always some points not in \(S\). Therefore, \(S\) is not path-connected. □

Definition (Open region in \(\mathbb{R}^n\))

A path-connected open set in \(\mathbb{R}^n\) is called an open region in \(\mathbb{R}^n\).

Example. \(\mathbb{R}^2\) is an open region. So is \(\{(x,y) | y > 0\}\). However, \(\{(x,y) | y \geq 0\}\) is not. □

Theorem (Any open ball is an open region). 

The \(r\)-neighbor of \(x=(x_1, x_2, \cdots, x_n)\) in \(\mathbb{R}^n\), \(N_r(x)\), is an open region.

Proof.   First, we show that \(N_r(x)\) is an open set. Take an arbitrary \(y \in N_r(x)\) and let \(d = d(x,y)\). Noting \(d < r\), let \(\delta = r - d (> 0)\). It suffices to show that \(N_{\delta}(y) \subset N_{r}(x)\). By the triangle inequality, for any \(z \in N_{\delta}(y)\), \[d(x,z) \leq d(x,y) + d(y,z) < d + \delta = r\] so that \(z \in N_{r}(x)\), and hence \(N_{\delta}(y) \subset N_{r}(x)\).

Next, we show that \(N_{r}(x)\) is path-connected. Let \(\sigma_y(t)\) denote a path (if exists) that starts at \(y\in N_{r}(x)\) (i.e., \(\sigma_y(0) = y\)) and ends at \(x\) (i.e., \(\sigma_y(1) = x\)). For any \(y, z \in N_{r}(x)\), define the following path:

\[\sigma_{yz}(t) = \left\{\begin{array}{lc} \sigma_{y}(2t) & \left(0 \leq t \leq \frac{1}{2}\right),\\ \sigma_{z}(2-2t) & \left(\frac{1}{2} < t \leq 1\right), \end{array} \right.\] then this is a path from \(y\) to \(z\) in \(N_{r}(x)\). We still need to show such path as \(\sigma_y(t)\) for each \(y = (y_1, y_2, \cdots, y_n) \in N_{r}(x)\) exists in \(N_{r}(x)\). Let us define, for example,

\[\begin{eqnarray} \sigma_y(t) &=& (tx_1 + (1-t)y_1, \cdots, tx_n + (1-t)y_n)\\ &=&tx + (1-t)y\\ &=&y + t(x - y) \end{eqnarray}\] for \(t \in [0, 1]\). This is a path (line segment) from \(y=\sigma_y(0)\) to \(x = \sigma_y(1)\). Furthermore, \[d(x,\sigma_y(t)) = \|\sigma_y(t) - x\| = (1-t)\|y - x\| = (1-t)d(x,y) < r.\] Thus, \(\sigma_y(t) \in N_r(x)\) for all \(t \in [0, 1]\). 

This proves that \(N_r(x)\) is path-connected. ■

Closed sets

The Intermediate Value Theorem can be extended to multivariate functions on a path-connected set. To extend the Extreme Value Theorem to multivariate functions, we need the notion of bounded closed sets.

Definition (Bounded set)

The subset \(S\) of \(\mathbb{R}^n\) is said to be bounded if there exist \(x \in \mathbb{R}^n\) and \(r > 0\) such that \(S \subset N_r(x)\).

That is, a set is bounded if it can be "covered" by an open ball.

Remark. Let \(d\) be the distance between \(x\) and the origin \(O(0, 0, \cdots, 0)\). Then (draw a figure!) \[N_r(x) \subset N_{d+r}(O).\] 

Therefore, in the above definition, we can replace the point \(x\) with the origin.

Example. In \(\mathbb{R}^2\), the rectangular region \([a, a']\times [b, b']\) where  \(-\infty < a < a' < \infty, -\infty < b < b' < \infty\) is bounded. To see this, let

\[r = \max\left\{\frac{a' - a}{2}, \frac{b' - b}{2}\right\}\] and \[P\left(\frac{a+a'}{2}, \frac{b+b'}{2}\right).\]

Then, for any real number \(R\) such that \(R > \sqrt{2}r\), we have \[[a, a']\times [b, b'] \subset N_R(P).\] □

Definition (Closed set)

The subset \(F\) of \(\mathbb{R}^n\) is said to be a closed set if \(\mathbb{R}^n\setminus F\) (i.e., the complement of \(F\) in \(\mathbb{R}^n\)), is an open set. A bounded closed set is a closed set that is bounded.

Remark. Be careful: "The set \(A\) is not an open set" does not necessarily imply that \(A\) is a closed set! See the following proposition.

Proposition

The semi-closed ("clopen") interval \([0, 1)\subset \mathbb{R}\) is neither open nor closed.

 Proof. Let \(x = 0\in [0, 1)\). For any \(\delta > 0\) and any point \(y \in (-\delta, 0) \subset N_{\delta}(0)\), we have \(y \not\in [0, 1)\). Hence \(N_\delta(0) \not\subset [0,1)\). Accordingly, \([0, 1)\) is not open. 

Next, consider the complement \([0, 1)^c = (-\infty, 0)\cup [1, \infty)\). Take \(x = 1 \in [0,1)^c\). For any \(\delta > 0\) and any point \(y \in (1 - \delta, 1) \subset N_{\delta}(1)\), we have \(y \not\in [0,1)^c\). Hence \(N_{\delta}(1) \not\subset [0,1)^c\). That is, \([0,1)^c\) is not open, and therefore, \([0, 1)\) is not closed. ■

Example. Let us show that the rectangular region \([a, a']\times [b, b']\) in \(\mathbb{R}^2\) is a closed set. Let \(F = [a, a']\times [b, b']\) and \(U = \mathbb{R}^2 \setminus F\). We need to show that \(U\) is open. Take an arbitrary point \(p = (x,y) \in U\). Since \(p \not\in F\), it is either \(x \not\in [a, a']\) or \(y \not\in [b, b']\). 

Consider the case that \(x \not\in [a, a']\) (the other case is similar). Then, either \(x < a\) or \(x > a'\). Consider the case that \(x > a'\) (the other case is similar). Let \(\delta = x - a'\). Then \(\delta > 0\). For any point \(q \in N_{\delta}(p)\), its \(x\)-coordinate is greater than \(a'\). Thus, \(q\not\in F\), that is, \(q \in U\). This implies that \(N_{\delta}(p) \subset U\). Hence \(U\) is open. Accordingly, \(F\) is closed. □