$\newcommand{R}{\mathbb{R}} \newcommand{N}{\mathbb{N}} \newcommand{defarrow}{\quad \stackrel{\text{def}}{\Longleftrightarrow} \quad}$

# Limits

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and $f \colon X \to Y$ a function between them.

## Sequential limits

A sequence $\{x_n\}_{n\in \N} \subset X$ is said to converge towards $x_0 \in X$ if for any $\varepsilon > 0$ there is a natural number $n_\varepsilon$ with the property that $x_n \in B_\varepsilon(x_0)$ for all $n \geq n_\varepsilon$: $\lim_{n \to \infty} x_n = x_0 \defarrow \forall\: \varepsilon > 0 \quad \exists\: n_\varepsilon \in \mathbb{N}; \qquad x_n \in B_{\varepsilon}(x_0) \quad\text{ for } \quad n \geq n_\varepsilon.$ We then say that $x_n$ tends to $x_0$ as $n$ tends to infinity, written $x_n \to x_0 \quad\text{(as }\quad n \to \infty), \qquad\text{ or }\qquad x_n \stackrel{n \to \infty}{\to} x_0.$ The point $x_0$ is called the limit of the sequence $\{x_n\}_{n \in \N}$.

### ℘ Sequential limits are zero limits for the distance function

Since $\{d_X(x_n,x_0)\}_{n\in \N}$ is a sequence in $\mathbb R$ it is easily verified that $x_n \to x_0 \quad\Longleftrightarrow\quad d_X(x_n,x_0) \to 0.$

Proof

Proof

\begin{align*} d_X(x_n,x_0) \to 0 &\quad\Longleftrightarrow\quad \forall \varepsilon > 0\quad \exists\, n_\varepsilon;\quad d_X(x_n,x_0) \in B_\varepsilon(0) \quad&(\text{in } \mathbb R) &\quad \text{ for } \quad n \geq n_\varepsilon\\ &\quad\Longleftrightarrow\quad \forall \varepsilon > 0\quad \exists \, n_\varepsilon;\quad d_X(x_n,x_0) < \varepsilon & &\quad \text{ for } \quad n \geq n_\varepsilon\\ &\quad\Longleftrightarrow\quad \forall \varepsilon >0 \quad \exists \, n_\varepsilon;\quad x_n \in B_\varepsilon(x_0) \quad& (\text{in } X) &\quad \text{ for } \quad n \geq n_\varepsilon\\ &\quad\Longleftrightarrow\quad x_n \to x_0. && \end{align*}

## Continuous limits (continuity)

We say that $f(x)$ converges to $y_0$ in $Y$ as $x$ converges to $x_0$ in $X$ if for any $\varepsilon > 0$ there exists $\delta > 0$ such that $f(x) \in B_\varepsilon(y_0)$ when $x \in B_\delta(x_0)$: $\lim_{x \to x_0} f(x) = y_0 \defarrow \forall\: \varepsilon > 0 \quad \exists\: \delta > 0; \qquad [d_X(x,x_0) < \delta \quad\Longrightarrow\quad d_Y(f(x),y_0) < \varepsilon].$ Equivalent ways of writing this are $f(x) \to y_0 \quad\text{ as }\quad x \to x_0, \qquad\text{ and }\qquad f(x) \stackrel{x \to x_0}{\to} y_0.$ A function $f$ satisfying this is said to be continuous at the point $x_0$. It is continuous on a set $D$ if it continuous at all points $x_0 \in D$, and simply continuous if its continuous on all of its domain.

Ex.
• The function $f\colon x \mapsto \frac{\sin(x)}{x}, \qquad x \in \R \setminus \{0\},$ may be extended to a bounded and continuous function $\R \to \R$, since $f(x) \stackrel{\text{in } \R}{\to} 1 \qquad\text{ as }\qquad x \stackrel{\text{in } \R}{\to} 0.$

### ℘ In metric spaces continuous and sequential limits agree

$(i) \quad \lim_{x \to x_0}f(x) = y \quad \Longleftrightarrow \quad (ii) \quad \lim_{n\to \infty} f(x_n) = y \text{ for any sequence such that } \lim_{n\to \infty} x_n = x_0.$

Proof

Proof

Assume that (i) holds, i.e., $d_Y(f(x),y) < \varepsilon \quad\text{ for }\quad d_X(x,x_0) < \delta.$ For any sequence $\{x_n\}_{n \in \N} \subset X$ with $\lim_{n\to \infty} x_n = x_0$, there exists $N \in \N$ with $d_X(x_n,x_0) < \delta \quad \text{ for }\quad n \geq N.$ Thus, according to (i), $d_Y(f(x_n),y) < \varepsilon \quad\text{ for }\quad n \geq N.$ This shows that (i) is sufficient for (ii) to hold.

Now assume that (i) does not hold, i.e., there is an $\varepsilon> 0$ such that for any $\delta > 0$ there exists $x_\delta$ with $d_X(x_\delta,x_0) < \delta \quad\text{ while }\quad d_Y(f(x_\delta),y) \geq \varepsilon.$ Thus, any sequence of $\delta_n \stackrel{n \to \infty}{\to} 0$ yields a sequence of numbers $x_n := x_{\delta_n}$ with $x_n \to x_0 \text{ as } n \to \infty \quad\text{ while }\quad d_Y(f(x_n),y) \geq \varepsilon.$ This violates (ii) and shows that (i) is necessary for (ii) to hold.

Ex.
• The sequence of functions given by $f_0(x) = 1, \quad f_1(x) = 1 - \frac{x^2}{3!}, \quad f_n(x) = \sum_{j=0}^n \frac{(-1)^j x^{2j}}{(2j+1)!} \quad n \in \N,$ converges in $BC([0,1],\R)$. Namely, let $f(x) = \frac{\sin(x)}{x}$, extended to a bounded and continuous function on $\R$ as in the preceding example. Since $\sin{x} \stackrel{\text{Taylor}}{=} x - \frac{x^3}{3!} + \ldots + \frac{(-1)^{n} x^{2n + 1}}{(2n+1)!} \pm \frac{\cos(\xi)\, x^{2n +3}}{(2n + 3)!}, \qquad 0 < |\xi| < |x|,$ we have $\frac{\sin{x}}{x} = \sum_{j=0}^n \frac{(-1)^j x^{2j}}{(2j+1)!} \pm \frac{\cos(\xi)\, x^{2n +2}}{(2n + 3)!}, \qquad 0 < |\xi| < |x|,$ so that \begin{align*} \|f_n - f\|_{BC([0,1],\R)} &= \sup_{x \in [0,1]} \bigg|\sum_{j=0}^n \frac{(-1)^j x^{2j}}{(2j+1)!} - \frac{\sin{x}}{x}\bigg|\\ &\leq \sup_{x,\xi \in [0,1]}\bigg| \frac{\cos(\xi)\, x^{2n +2}}{(2n + 3)!} \bigg| = \frac{1}{(2n+3)!} \to 0 \quad\text{ as }\quad n \to \infty.\end{align*}

### ℘ Limits are unique

If $\lim_{n \to \infty} x_n = x$ and $\lim_{n \to \infty} x_n = y$, then $x = y$.

Proof

Proof

In view of the assumptions, and using the triangle inequality, $0 \leq d(x,y) \leq d(x,x_n) + d(x_n,y) \to 0 \quad \text{ as } \quad n \to \infty.$ Then $d(x,y) = 0$ implies $x=y$ by the axioms of a metric space.

Ex.
• The sequence $x_1 = 0.9, \quad x_2 = 0.99, \quad x_3 = 0.999, \quad\text{ and so forth,}$ converges towards $0.999\ldots$ in $\R$, but also towards $1$. Hence $0.999\ldots = 1.$

## Accumulation points

A concept related to convergence is that of an accumulation point of a subset $M \subset X$: $x_0 \text{ accumulation point for } M \defarrow \exists \{x_n\}_{n\in \N} \subset M; \quad x_n \to x_0.$ Equivalently, any small ball $B_\varepsilon(x_0)$ centered at $x_0$ contains a point in $M$.

N.b. The limit of a sequence is always an accumulation point for that sequence, but a (non-convergent) sequence may have several, or no, accumulation points. An accumulation point for a sequence is, per definition, the limit of a subsequence of that sequence.

Ex.
• $0$ is an accumulation point for $\{1/n\}_{n \in \N}$, but also for $\{1,1,2,\frac{1}{2},3,\frac{1}{3},4,\frac{1}{4}, \ldots\}.$

## Relationship between limits and closures

### ℘ Closures are the total of sequential limits of interior points

By comparing the definitions of boundary and interior points with that of a sequential limit, one obtains that $\mathrm{clos} (D;X) = \{ x \in X \colon x = \lim_{n \to \infty} x_n \text{ for some sequence } \{x_n\}_{n\in \N} \subset D\}$

Proof

Proof

Assume that $x \in \overline D$. Then, according to the definitions of interior and boundary points, any small ball $B_{1/n}(x)$ contains a point $x_n \in D$. This means that $\{x_n\}_{n\in \N}\subset D$ converges to $x$.

Now, assume instead that there is a sequence $\{x_n\}_{n\in \N} \subset D \quad\text{ with }\quad \lim_{n \to \infty} x_n = x \quad\text{ in } X.$ Then $d_X(x_n,x) \to 0$ as $n \to \infty$, so that $\forall\: \varepsilon > 0 \quad \exists\: x_{n_\varepsilon} \in B_\varepsilon(x).$ Since $x_{n_\varepsilon} \in D$, either $x$ is an interior point (for small $\varepsilon$ there are only points from $D$ in $B_\varepsilon(x)$), or $x$ is a boundary point ($B_\varepsilon(x)$ contains also points from the complement of $D$); in any case $x \in \overline{D}$.

Ex.
• In $l_\infty$, $\textstyle{(\frac{1}{2},\frac{2}{3},\frac{3}{4},\ldots)} \in \overline{B_1},$ since it is the limit of the sequence $x_1 = (\frac{1}{2}, \frac{1}{2}, \ldots)$, $x_2 = (\frac{1}{2}, \frac{2}{3}, \frac{2}{3}, \ldots)$, $x_3 = (\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{3}{4}, \ldots)$, and so forth (the elements of which are all in $B_1$ in $l_\infty$). To see this, let $x_0 := \{\frac{j}{j+1}\}_{j \geq 1}$. Then $\|x_0 -x_n\|_{l_\infty} = \sup_{j\geq n} \Big|\frac{j}{j+1} - \frac{n}{n+1}\Big| = \Big| 1 - \frac{n}{n+1}\Big| = \frac{1}{n+1} \to 0 \quad \text{ as } \quad n \to \infty,$ in view of that the function $j \mapsto \frac{j}{j+1}$ is monotone increasing and bounded by $1$. Thus $B_1 \ni x_n \stackrel{\text{in } l_\infty}{\to} \textstyle{(\frac{1}{2},\frac{2}{3},\frac{3}{4},\ldots)}$ as $n \to \infty.$