Let $X$ be a locally compact Hausdorff space. Then $X$ is a Baire space.

Let $(X, \tau)$ be a second-countable topological space. Then $X$ is Lindelof.

Let $\{X_\alpha\}_{\alpha \in A}$ be a family of topological spaces with $|A| \ge 2$, and let each $X_\alpha$ be nonempty. The product $\prod_{\alpha \in A} X_\alpha$ (with the product topology) is locally compact if and only if each $X_\alpha$ is locally compact and all but finitely many of the $X_\alpha$ are compact.

Let $X$ be a locally compact Hausdorff space. Then $C_0(X)$, equipped with the supremum norm, is a Banach space. Moreover, $C_c(X)$ is dense in $C_0(X)$.

Let $1 \le p < \infty$ and let $\mathcal{F} \subset L^p(\mathbb{R}^n)$. Then $\mathcal{F}$ is relatively compact in $L^p(\mathbb{R}^n)$ if and only if: 1. $\mathcal{F}$ is **bounded**: $\sup_{f \in \mathcal{F}} \|f\|_{L^p} < \infty$. 2. $\mathcal{F}$ is **$L^p$-equicontinuous in translation**: $\lim_{|h| \to 0} \sup_{f \in \mathcal{F}} \|f(\cdot + h) - f(\cdot)\|_{L^p} = 0$. 3. $\mathcal{F}$ is **tight**: $\lim_{R \to \infty} \sup_{f \in \mathcal{F}} \int_{|x| > R} |f(x)|^p \, d\mathcal{L}^n = 0$.

Let $K$ be a compact metric space, and let $f_n: K \to \mathbb{R}$ be a sequence of functions that is equicontinuous on $K$. If $f_n \to f$ pointwise on $K$, then: 1. The limit $f$ is continuous on $K$. 2. The convergence $f_n \to f$ is uniform on $K$.

Let $U \subset \mathbb{R}^n$ be open and convex, and let $\mathcal{F} \subset C^1(U; \mathbb{R})$ be a family of continuously differentiable functions. If there exists $M > 0$ such that \begin{align*} |\nabla f(x)| \le M \quad \text{for all } f \in \mathcal{F} \text{ and all } x \in U, \end{align*} then $\mathcal{F}$ is uniformly equicontinuous on $U$ with modulus $\delta = \varepsilon / M$.

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces, and let $\mathcal{F}$ be a family of functions from $X$ to $Y$. Suppose there exists a constant $L \ge 0$ such that every $f \in \mathcal{F}$ satisfies \begin{align*} d_Y(f(x), f(y)) \le L \cdot d_X(x, y) \quad \text{for all } x, y \in X. \end{align*} Then $\mathcal{F}$ is uniformly equicontinuous with modulus $\delta = \varepsilon / L$ (for $L > 0$).

For any subset $A$ of a metric space $(M, d)$ and any $\varepsilon > 0$: \begin{align*} \mathcal{P}(A, 2\varepsilon) \le \mathcal{N}(A, \varepsilon) \le \mathcal{P}(A, \varepsilon). \end{align*}

Let $(K, d_K)$ be a compact metric space, and let $\mathcal{F} \subset C(K)$. Then $\mathcal{F}$ is totally bounded in $(C(K), \|\cdot\|_\infty)$ if and only if: 1. $\mathcal{F}$ is **uniformly bounded**: there exists $M > 0$ with $\|f\|_\infty \le M$ for all $f \in \mathcal{F}$. 2. $\mathcal{F}$ is **equicontinuous**: for every $\varepsilon > 0$, there exists $\delta > 0$ such that $d_K(x, y) < \delta$ implies $|f(x) - f(y)| < \varepsilon$ for all $f \in \mathcal{F}$ and all $x, y \in K$.