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

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

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

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

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

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

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 \(n \to \infty.\)