Continuity

Let $X \subseteq \mathbb{R}$ be a subset of real numbers, $x_0 \in X$ and $f: X \rightarrow \mathbb{R}$ be a function that maps elements of $X$ to real numbers. The function $f$ is continuous at $x_0$ if and only if the limit of $f(x)$ as $x \rightarrow x_0$ is equal to $f(x_0).$

$$\lim_{x \rightarrow x_0} f(x) = f(x_0)$$

This uses the limit of a function, which requires $x_0$ to be a limit point of $X.$ If $x_0$ is an isolated point of $X$ (some open interval around $x_0$ contains no other point of $X$) then $f$ is continuous at $x_0$ by convention. Written out, $f$ is continuous at $x_0$ if and only if for every $\varepsilon>0$ there exists $\delta>0,$ which may depend on both $\varepsilon$ and $x_0,$ such that

$$\lvert f(x)-f(x_0)\rvert\leq\varepsilon\,\,\forall\,x\in X,\,\lvert x-x_0\rvert<\delta.$$

The function $f$ is continuous on $X$ if and only if $f$ is continuous at $x \, \, \forall \, x \in X.$ Otherwise, $f$ is discontinuous.

Let $f,g:X\rightarrow\mathbb{R}$ be continuous at $x_0\in X$ and $c\in\mathbb{R}.$ The limit laws give that $f+g,$ $f-g,$ $c\,f$ and $f\,g$ are continuous at $x_0,$ and $\frac{f}{g}$ is continuous at $x_0$ if $g(x_0)\neq 0.$ Let $Y\subseteq\mathbb{R},$ $f:X\rightarrow Y$ be continuous at $x_0$ and $g:Y\rightarrow\mathbb{R}$ be continuous at $f(x_0).$ The composition $g\circ f:X\rightarrow\mathbb{R}$ is continuous at $x_0.$ Constant functions and the identity function $x\mapsto x$ are continuous, so every polynomial is continuous on $\mathbb{R}$ and every rational function is continuous wherever its denominator is non-zero.

Let $f_n:X\rightarrow\mathbb{R}$ be a sequence of functions that converges uniformly to $f:X\rightarrow\mathbb{R}.$ If every $f_n$ is continuous at $x_0\in X$ then $f$ is continuous at $x_0,$ so a uniform limit of continuous functions is continuous. Pointwise convergence does not suffice, since $f_n(x)=x^n$ converges pointwise on $\left[0,1\right]$ to a discontinuous function. A power series converges uniformly on every closed interval inside its interval of convergence and its partial sums are polynomials, so it is continuous on $\left(c-R,c+R\right);$ in particular the exponential, cosine and sine functions are continuous on $\mathbb{R}.$

The intermediate value theorem states that if a function $f:[a,b]\rightarrow\mathbb{R}$ is continuous and $C$ lies strictly between $f(a)$ and $f(b)$ (either $f(a)<C<f(b)$ or $f(b)<C<f(a)$) then there is at least one $c\in(a,b) : f(c)=C.$

Let $f:X\rightarrow\mathbb{R}$ be a function. $f$ is bounded on $X$ if and only if its image $f(X)$ is a bounded set.

$$\exists M\geq 0:\lvert f(x)\rvert\leq M\,\,\forall\,x\in X$$

The extreme value theorem states that if $f:[a,b]\rightarrow\mathbb{R}$ is continuous on a closed bounded interval then $f$ is bounded and attains its maximum and minimum.

$$\exists\,x_{\min},x_{\max}\in[a,b]:f\left(x_{\min}\right)\leq f(x)\leq f\left(x_{\max}\right)\,\,\forall\,x\in[a,b]$$

Both the intermediate and extreme value theorems can fail on intervals that are not closed and bounded. For example, $f(x)=\frac{1}{x}$ is continuous on $(0,1]$ but unbounded.

Let $X$ be an interval and $f:X\rightarrow\mathbb{R}$ be continuous. The image $f(X)$ is an interval, since by the intermediate value theorem it contains every value between any two of its elements. If $X=[a,b]$ is closed and bounded then, by the extreme value theorem, $f([a,b])$ is the closed bounded interval $\left[f\left(x_{\min}\right),f\left(x_{\max}\right)\right].$

A function $f:X\rightarrow\mathbb{R}$ is uniformly continuous on $X$ if and only if for every $\varepsilon>0$ there exists $\delta>0$ that depends on $\varepsilon$ alone such that $f(x)$ and $f(y)$ are $\varepsilon$-close whenever $x,y\in X$ are within $\delta$ of each other.

$$\forall\varepsilon>0\,\exists\delta>0:\lvert f(x)-f(y)\rvert\leq\varepsilon\,\,\forall\,x,y\in X,\,\lvert x-y\rvert<\delta$$

A uniformly continuous function is continuous. The converse is false: $f(x)=x^2$ is continuous but not uniformly continuous on $\mathbb{R},$ and $f(x)=\frac{1}{x}$ is continuous but not uniformly continuous on $(0,1].$ The Heine–Cantor theorem states that a continuous function on a closed bounded interval $[a,b]$ is uniformly continuous.

A function $f:X\rightarrow\mathbb{R}$ is Lipschitz continuous on $X$ if and only if there exists a Lipschitz constant $L\geq 0$ that bounds how fast $f$ can change.

$$\lvert f(x)-f(y)\rvert\leq L\,\lvert x-y\rvert\,\,\forall\,x,y\in X$$

A Lipschitz continuous function is uniformly continuous, since $\delta=\frac{\varepsilon}{L}$ works for every $\varepsilon>0$ when $L>0.$ The converse is false: $f(x)=\sqrt{x}$ is uniformly continuous but not Lipschitz continuous on $[0,1].$

Let $X$ be an interval and $f:X\rightarrow\mathbb{R}$ be a function. $f$ is monotonically increasing if and only if $f(x)\leq f(y)$ whenever $x<y$ and strictly increasing if and only if $f(x)<f(y)$ whenever $x<y.$ $f$ is monotonically decreasing or strictly decreasing if and only if $-f$ is monotonically or strictly increasing respectively. $f$ is monotonic if it is either monotonically increasing or monotonically decreasing, and strictly monotonic if it is either strictly increasing or strictly decreasing.

A monotonic function $f$ on an interval $X$ is continuous if and only if its image $f(X)$ is an interval. A monotonic function can be discontinuous at no more than countably many points of $X.$

Let $X$ be an interval and $f:X\rightarrow\mathbb{R}$ be continuous and strictly monotonic. Then $f$ is injective, its image $Y=f(X)$ is an interval and its inverse function $f^{-1}:Y\rightarrow X$ is continuous and strictly monotonic in the same direction as $f.$ Conversely, a continuous injective function on an interval is strictly monotonic. For example, $f(x)=x^n$ is continuous and strictly increasing on $[0,\infty)$ for $n\in\mathbb{N},$ so the $n$-th root $f^{-1}(y)=y^{\frac{1}{n}}$ is continuous and strictly increasing on $[0,\infty).$