Let $X$ be a set equipped with the discrete topology $\tau_X = \mathcal{P}(X)$, let $(Y,\tau_Y)$ be a topological space, and let \begin{align*} f: X &\to Y \end{align*} be any map. Then $f: (X,\tau_X) \to (Y,\tau_Y)$ is continuous.

Let $(X,\tau)$ be a [topological space](/page/Topological%20Space), and let $A \subset X$. Define the [interior](/page/Interior) of $A$ by \begin{align*} A^\circ := \bigcup \{V \subset X : V \in \tau \text{ and } V \subset A\}. \end{align*} Then $A^\circ \in \tau$, $A^\circ \subset A$, and for every $U \in \tau$ such that $U \subset A$, one has $U \subset A^\circ$.

Let $(X,\tau_X)$, $(Y,\tau_Y)$, and $(Z,\tau_Z)$ be topological spaces. Let $f: X \to Y$ and $g: Y \to Z$ be continuous maps, where continuity means that the preimage of every open set is open in the domain topology. Then the composition \begin{align*} g \circ f: X &\to Z \\ x &\mapsto g(f(x)) \end{align*} is continuous.

Let $(X,\tau)$ be a topological space, where $\tau$ is the collection of open subsets of $X$. A subset $F \subset X$ is closed if its complement $X \setminus F$ belongs to $\tau$. If $(F_i)_{i\in I}$ is any indexed family of closed subsets of $X$, then $\bigcap_{i\in I} F_i$ is closed. If $n \in \mathbb{N}$ and $F_1,\ldots,F_n$ are closed subsets of $X$, then $\bigcup_{i=1}^n F_i$ is closed.

Let $X$ be a set, and let $\mathcal B \subset \mathcal P(X)$ be a basis for a topology on $X$, meaning that $\bigcup_{B \in \mathcal B} B = X$ and whenever $B_1, B_2 \in \mathcal B$ and $x \in B_1 \cap B_2$, there exists $B_3 \in \mathcal B$ such that $x \in B_3 \subset B_1 \cap B_2$. Define \begin{align*} \tau_{\mathcal B} := \{U \subset X : \text{for every } x \in U, \text{ there exists } B \in \mathcal B \text{ such that } x \in B \subset U\}. \end{align*} Then $\tau_{\mathcal B}$ is a topology on $X$, and every element of $\tau_{\mathcal B}$ is a union of elements of $\mathcal B$.

Let $(X,\tau)$ be a [topological space](/page/Topological%20Space), and let $A \subset X$. Then the closure $\overline{A}$ is closed in $(X,\tau)$, satisfies $A \subset \overline{A}$, and has the following minimality property: for every subset $F \subset X$, if $F$ is closed in $(X,\tau)$ and $A \subset F$, then $\overline{A} \subset F$.

Let $\Omega\subset\mathbb C$ be open, let $f:\Omega\to\mathbb C$ be holomorphic, and let $z_0\in\Omega$ satisfy $f'(z_0)\ne0$. Then there exist open neighbourhoods $U\subset\Omega$ of $z_0$ and $V\subset\mathbb C$ of $f(z_0)$ such that $f|_U:U\to V$ is bijective, $(f|_U)^{-1}:V\to U$ is holomorphic, and \begin{align*} \left((f|_U)^{-1}\right)'(w)=\frac{1}{f'((f|_U)^{-1}(w))} \end{align*} for every $w\in V$.

Let $\Omega \subset \mathbb{C}$ be an open set, and let $f: \Omega \to \mathbb{C}$ and $g: \Omega \to \mathbb{C}$ be holomorphic functions. Then the functions $f+g: \Omega \to \mathbb{C}$, $fg: \Omega \to \mathbb{C}$, and $af: \Omega \to \mathbb{C}$ are holomorphic on $\Omega$ for every scalar $a \in \mathbb{C}$. If $g(z) \ne 0$ for every $z \in \Omega$, then the quotient function $f/g: \Omega \to \mathbb{C}$ is holomorphic on $\Omega$.

Let $U\subsetneq\mathbb R^n$ be open and let $L$ be a second-order nondivergence form operator with real symmetric coefficient matrix $A=(a_{ij})$. Then $A$ is uniformly elliptic with constant $\theta>0$ iff \begin{align*} p_L(x,\xi)&\ge \theta |\xi|^2 \end{align*} for $\mathcal L^n$-a.e. $x\in U$ and every $\xi\in\mathbb R^n$.

Let $U\subsetneq\mathbb R^n$ be bounded and open. Let $B:H_0^1(U)\times H_0^1(U)\to\mathbb R$ be bounded and coercive. For every $F\in H^{-1}(U)$, there exists a unique $u\in H_0^1(U)$ such that \begin{align*} B[u,v]&=F(v) \end{align*} for every $v\in H_0^1(U)$.