# Useful limit theorem

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22.03.2012

Here is a useful limit theorem which can be used to prove many of the commonly needed limit theorems as special cases.

## Claim

Let $(X^*, T_{X^*})$, $(X,T_X)$, $(Y, T_Y)$, and $(Z, T_Z)$ be topological spaces, where $X$ is a subspace of $X^*$. Let $f : X \to Y$ be such that $\lim_{x \to p} f(x) = c$, where $c \in Y$ and $p \in \overline{X}$. Let $g : Y \to Z$ be continuous at $c$. Then

## Example 1 : The limit of evaluations is the evaluation of the limit

Let $f : X \to X$ be such that $f(x) = x$, and $g : X \to Z$ be continuous at $p \in X$. Then

Actually, this is equivalent to continuity at $p$.

## Example 2: The limit of sequential evaluations is the evaluation of the limit

Let $f : \mathbb{N} \to Y$, where $\mathbb{N}$ is a subspace of $\mathbb{N}^* = \mathbb{N} \cup \{\infty\}$. Let $g:Y \to Z$ be a continuous function. Then

## Example 3 : The limit of a component is the component of a limit

Let $f : X \to Z^n$ be such that $f(x) = (f_1(x), \dots, f_n(x))$ and $\pi_i : Z^n \to Z$ be such that $\pi_i(z) = z_i$. Since the projection $\pi_i$ is continuous,

Actually, the reverse implication also holds: if the limits of all component functions exist, then so does the limit of the function (with the relation above).

## Example 4 : Semigroup addition

Let $X$ and $Z$ be topological semigroups (e.g vector spaces), $f : X \to Z^2$, and $g : Z^2 \to Z$ be such that $g(x,y) = x + y$. Since addition is continuous,

We also used the result of example 3 here.

## Example 5 : Module multiplication

Let $X$, $Y$, and $Z$ be topological $R$-modules (e.g. vector spaces), $f : X \to Y$, and $g : Y \to Z$ be such that $g(y) = \alpha y$, where $\alpha \in \mathbb{R}$. Since by definition the multiplication is continuous, so is its restriction to a fixed $\alpha$, and thus

## Proof

Denote by $T_Z(z)$ the neighborhoods of a point $z \in Z$ (and similarly for other spaces). Let $c = \lim_{x \to p} f(x)$, and $W \in T_Z(g(c))$. Since $g$ is continuous at $c$, there exists $V \in T_Y(c)$ such that $g(V) \subset W$. By the definition of a limit of a function, there exists $U \subset T_{X^*}(p)$ such that $f(X \cap U \setminus \{p\}) \subset V$. Therefore $g(f(X \cap U \setminus \{p\})) \subset W$, i.e. $\lim_{x \to p} g(f(x)) = g(c)$. QED.