Heine theorem

Not to be confused with Heine–Borel theorem.

Heine's theorem, named after the German mathematician Eduard Heine, establishes a link in mathematical analysis between limits of functions and limits of sequences. The theorem states that the existence and value of the limit of a function at a point can be characterized by the limits of all sequences that converge to that point. Conversely, information about sequential limits can be used to determine function limits. As a consequence, many properties of limits of functions may be derived from the corresponding properties of limits of sequences.

Specifically, it contains a statement with two parts:

Forward statement: Let $f$ be a function and let $a$ be an accumulation point of its domain. If $$\lim \limits _{x\to a}f\left(x\right)=L,$$ then for every sequence $\left\{x_{n}\right\}$ that converges to $a$ and satisfies $x_{n}\neq a$ for all $n$, the sequence $\left\{f\left(x_{n}\right)\right\}$ converges to $L$; that is, $$\lim \limits _{x\to a}f\left(x\right)=L\implies \left(\forall \left\{x_{n}\right\}:\ x_{n}\to a,\forall n,x_{n}\neq a\implies \left\{f\left(x_{n}\right)\right\}\to L\right).$$

Converse statement: Conversely, if for every sequence $\left\{x_{n}\right\}$ with $x_{n}\to a$ and $x_{n}\neq a$ for all $n$, the sequence $\left\{f(x_{n})\right\}$ converges, and all such sequences have the same limit $L$, then the limit of $f$ at $a$ exists and equals $L$; in symbols, $$\left(\forall \left\{x_{n}\right\}:\ x_{n}\to a,\forall n,x_{n}\neq a\implies \left\{f\left(x_{n}\right)\right\}\to L\right)\implies \lim \limits _{x\to a}f\left(x\right)=L.$$

Background

In several contexts, the topology of a space is conveniently specified in terms of limit points. This is often accomplished by specifying when a point is the limit of a sequence. Still, for some spaces that are too large in some sense, one specifies also when a point is the limit of more general sets of points indexed by a directed set, known as nets. A function is (Heine-)continuous only if it takes limits of sequences to limits of sequences. In the former case, preservation of limits is also sufficient; in the latter, a function may preserve all limits of sequences yet still fail to be continuous, and preservation of nets is a necessary and sufficient condition.

In detail, a function $f:X\to Y$ is sequentially continuous if whenever a sequence $\left(x_{n}\right)$ in $X$ converges to a limit $x,$ the sequence $\left(f\left(x_{n}\right)\right)$ converges to $f(x).$ Thus, sequentially continuous functions "preserve sequential limits." Every continuous function is sequentially continuous. If $X$ is a first-countable space and countable choice holds, then the converse also holds: any function preserving sequential limits is continuous. In particular, if $X$ is a metric space, sequential continuity and continuity are equivalent. For non-first-countable spaces, sequential continuity might be strictly weaker than continuity. (The spaces for which the two properties are equivalent are called sequential spaces.) This motivates the consideration of nets instead of sequences in general topological spaces. Continuous functions preserve the limits of nets, and this property characterizes continuous functions.

Formal statement

Consider the case of real-valued functions of one real variable:

Theorem—A function $f:A\subseteq \mathbb {R} \to \mathbb {R}$ is continuous at $x_{0}$ if and only if it is sequentially continuous at that point.

Proof

Proof. Assume that $f:A\subseteq \mathbb {R} \to \mathbb {R}$ is continuous at $x_{0}$ (in the sense of $\epsilon -\delta$ continuity). Let $\left(x_{n}\right)_{n\geq 1}$ be a sequence converging at $x_{0}$ (such a sequence always exists, for example, $x_{n}=x,{\text{ for all }}n$); since $f$ is continuous at $x_{0}$ $$\forall \epsilon >0\,\exists \delta _{\epsilon }>0:0<|x-x_{0}|<\delta _{\epsilon }\implies |f(x)-f(x_{0})|<\epsilon .\quad (*)$$ For any such $\delta _{\epsilon }$ we can find a natural number $\nu _{\epsilon }>0$ such that for all $n>\nu _{\epsilon },$ $$|x_{n}-x_{0}|<\delta _{\epsilon },$$ since $\left(x_{n}\right)$ converges at $x_{0}$; combining this with $(*)$ we obtain $$\forall \epsilon >0\,\exists \nu _{\epsilon }>0:\forall n>\nu _{\epsilon }\quad |f(x_{n})-f(x_{0})|<\epsilon .$$ Assume on the contrary that $f$ is sequentially continuous and proceed by contradiction: suppose $f$ is not continuous at $x_{0}$ $$\exists \epsilon >0:\forall \delta _{\epsilon }>0,\,\exists x_{\delta _{\epsilon }}:0<|x_{\delta _{\epsilon }}-x_{0}|<\delta _{\epsilon }\implies |f(x_{\delta _{\epsilon }})-f(x_{0})|>\epsilon$$ then we can take $\delta _{\epsilon }=1/n,\,\forall n>0$ and call the corresponding point $x_{\delta _{\epsilon }}=:x_{n}$: in this way we have defined a sequence $(x_{n})_{n\geq 1}$ such that $$\forall n>0\quad |x_{n}-x_{0}|<{\frac {1}{n}},\quad |f(x_{n})-f(x_{0})|>\epsilon$$ by construction $x_{n}\to x_{0}$ but $f(x_{n})\not \to f(x_{0})$, which contradicts the hypothesis of sequential continuity. $\blacksquare$