\( \newcommand\R{\mathbb{R}} \newcommand\Q{\mathbb{Q}} \newcommand\Z{\mathbb{Z}} \newcommand\N{\mathbb{N}} \newcommand\ve[1]{\boldsymbol{#1}} \newcommand\norm[1]{\left\Vert #1 \right\Vert} \def\u{\ve{u}} \def\v{\ve{v}} \def\w{\ve{w}} \)
We define open and closed intervals as follows. \[ (a, b) = \{x \in \R: a < x < b\}, \] \[ [a, b] = \{x \in \R: a \leq x \leq b\}. \]
An \(n\)-cell is the Cartesian product of \(n\) closed intervals.
A metric space is a set \(M\) with a metric or a distance function \(d\colon M\times M \to \R\) such that for all \(x, y, z \in M\),
An open ball of radius \(r\) in a metric space \((M, d)\) centred at a point \(x\) is the set of all \(y\) such that \(d(x, y) < r\). This is denoted as \(B_r(x)\).
A closed ball of radius \(r\) in a metric space \((M, d)\) centred at a point \(x\) is the set of all \(y\) such that \(d(x, y) \leq r\).
An open ball \(B_r(x)\) is also called a neighbourhood \(N_r(x)\) of \(x\).
A convex set \(E\) is such that for any \(x, y \in E\), \[ tx + (1 - t)y \in E, \] for all \(t \in [0, 1]\). Open balls, closed balls, and \(n\)-cells are all convex sets.
A bounded set \(S \subseteq M\) is such that there exists \(x \in M\) and some radius \(r > 0\) where \(S \subseteq B_r(x)\).
A point \(\ell \in M\) is a closure point of \(S \subseteq \R\) if every neighbourhood of \(\ell\) contains at least one point of \(S\). The set of all closure points of \(S\) is called its closure, denoted by \(\bar{S}\). A closed set is such that \(S = \bar{S}\).
A subset \(S\) of a metric space is dense in \(M\) if \(\bar{S} = M\).
A limit point \(\ell\) of \(S \subseteq M\) is such that every neighbourhood of \(\ell\) contains at least one point of \(S\) different from itself. Every closure point of \(S\) which is not a limit point of \(S\) is called an isolated point.
Note that every neighbourhood of a limit point of \(S\) contains infinitely many points of \(S\). Suppose \(\ell\) is a limit point of \(S\), and suppose that the neighbourhood \(N\) of \(\ell\) contains finitely many points in \(S\) apart from \(\ell\). Thus, we can enumerate \((N\cap S)\setminus\{\ell\} = \{s_1, \dots, s_n\}\). Note that if \(d(s, \ell) > d(s_i, \ell)\), the neighbourhood \(N_{r_i}(\ell)\) does not contain \(s\), where \(r_i = d(s_i, \ell)\). Thus, by setting \[ r = \min d(s_i, \ell) > 0,\] the neighbourhood \(N_r(\ell)\) contains none of \(s_i\). This is a contradiction.
As a corollary, finite subsets of a metric space have no limit points.
An interior point \(p \in S\) is such that for some \(r > 0\), there is a neighbourhood \(B_r(p) \subseteq S\). The set of all interior points of \(S\) is called its interior, denoted by \(S^0\). An open set is such that \(S = S^0\).
Note that every neighbourhood is an open set.
The boundary \(\partial S\) of \(S\) is the set of all points in its closure not in its interior. Thus, \(\partial S = \bar{S} \setminus S^0\).
A set \(S\) is open iff its complement is closed. As a corollary, a set is closed iff its complement is open.
Let \(S\) be a set and let \(E_i\) be subsets of \(S\). Then, \[ \left(\bigcup E_i \right)^c = \bigcap E_i^c\, \quad\quad \left(\bigcap E_i \right)^c = \bigcup E_i^c. \]
Let \(\{\mathcal{O}_i\}\) be a collection of open sets. Then, \(\cup_i \mathcal{O}_i\) is open. It immediately follows that the complement \(\cap_i \mathcal{O}_i^c\) is closed.
Let \(\{\mathcal{C}_i\}\) be a collection of closed sets. Then, \(\cap_i \mathcal{C}_i\) is closed. Note that this follows since the complement \(\cup_i \mathcal{C}_i^c\) is open (remember that \(\mathcal{C}_i^c\) are all open).
Let \(\{\mathcal{O}_i\}\) be a finite collection of open sets. Then, \(\cap_i \mathcal{O}_i\) is open.
For infinitely many open sets, consider \(\mathcal{O}_n = (-1/n, 1/n)\) for \(n \in \N\).
Let \(\{\mathcal{C}_i\}\) be a finite collection of closed sets. Then, \(\cup_i \mathcal{C}_i\) is closed.
Let \(S\subseteq\R\) be closed. Then, if \(S\) is bounded above, it has a maximum. If it is bounded below, it has a minimum.
Let \((M, d)\) be a metric space and let \(S \subset M\). Then \(T \subseteq S\) is open in \(S\) iff \(T = S \cap A\) is open for some open subset \(A\) of \(M\).
An open cover of \(S \subseteq M\) is a collection of open subsets of \(M\) whose union is \(S\).
A subset \(K \subseteq M\) is called compact if every open cover of \(K\) contains a finite subcover. For example, any finite set is compact.
Let \(S \subset M\). A set \(K \subseteq S\) is compact in \(S\) iff \(K\) is compact in \(M\).
Recall that the same does not hold for openness. For example, \(\R\) is open in \(\R\), but not in \(\R^2\).
All compact subsets of a metric space are closed. Suppose \(K \subseteq M\) is compact. If \(K\) is not closed, there is some \(x \in K^c\) such that \(x\) is a limit point of \(K\). Construct the open cover of \(K\) with open balls \(B(k, r_k)\) centred at \(k\), of radius \(r_k = d(x, k)/2\) for all \(k \in K\). Note that an open ball \(B(x, r_k)\) of the same radius centred at \(x\) is disjoint from the first ball. From the compactness of \(K\), the cover \(\{B(k, r_k)\}_{k \in K}\) has a finite subcover. Now take the intersection of all the open balls \(B(x, r_k)\). This is a neighbourhood of \(x\) which contains no element of \(K\), which is a contradiction.
All closed subsets of compact spaces are compact.
The intersection of a closed set and a compact set is compact.
For any collection of compact subsets of \(\{K_j\}\) such that the intersection of any finite sub-collection is nonempty, the intersection of all of them is also nonempty.
Given a collection of closed intervals \(\{I_j\}\) in \(\R\) such that \(I_{n + 1} \subseteq I_n\) for all \(n \in \N\), the intersection of all of them is nonempty. The same holds even if \(\{I_j\}\) are a collection of \(n\)-cells.
Every infinite subset of a compact set \(K\) has a limit point in \(K\).
Every \(n\)-cell is compact.
Every bounded infinite subset of \(\R^n\) has a limit point in \(\R^n\).
To prove this, given an infinite bounded subset \(S \subseteq \R^n\), we construct an \(n\)-cell \(I\) around it, so \(S \subset I\). Now, \(I\) is compact, so \(S\) must have a limit point in \(I\), hence a limit point in \(\R^n\).
If \(\{S_m\}_{m \in \N}\) is a collection of bounded nonempty closed sets such that \(S_{m + 1} \subseteq S_m\) for all \(m \in \N\), then \(\bigcap_{m \in \N} S_m \neq \emptyset\).
To prove this, note that either all \(S_m\) are infinite, or some \(S_{m_0}\) is finite. If the latter is true, then in order for all \(S_{m > m_0}\) to be nonempty, there must be some \(S_{n > m_0} \neq \emptyset\) such that all subsequent sets \(S_{m \geq n} = S_n\). This is because the number of elements in each \(S_{m > m_0}\) must decrease or stay constant, but never reach zero. Thus, the intersection of all \(S_m\) is simply \(S_n\), which is nonempty.
In the case where all \(S_m\) are infinite, choose \(A = \{a_1, a_2, \dots\}\) where \(a_i \in S_i\) are distinct for all \(i \in \N\). Since \(A \subseteq S_1\), it is bounded. Thus, by the Bolzano-Weierstrass theorem, we find a limit point \(\ell\) of \(A\) in \(\R\). Since \(A\setminus(A \cap S_n)\) is finite, it follows that \(\ell\) is a limit point of \(A \cap S_n\). Thus, \(\ell\) is a limit point of the closed set \(S_n\), for all \(n \in \N\). Hence, \(\ell \in S_n\) for all \(n \in \N\).
For \(S \subset \R^n\), the following are equivalent.