Section 5.1
A bit of analysis
Given two real numbers, we can compare them, and so there is a notion of smallness.
Let $\lbrace a_n\rbrace _{n \ge 0}$ be an infinite sequence of real numbers. We say that its limit is $L$ if for every $\epsilon > 0$, there exists a sufficiently large $N_\epsilon > 0$ such that
$$ \lvert a_n - L \rvert < \epsilon $$
for all $n \ge N_\epsilon$. In this case, we write
$$ \lim_{n \to \infty} a_n = L. $$
Of course, some sequences do not have limits. For instance, $a_n = (-1)^n$ does not have a limit. But if a limit exists, it is unique.
Show that if $\lim_{n \to \infty} a_n = a$ and $\lim_{n \to \infty} b_n = b$, then $\lim_{n \to \infty} (a_n + b_n) = a + b$.
Show that if $\lim_{n \to \infty} a_n = a$ and $\lim_{n \to \infty} b_n = b$, then $\lim_{n \to \infty} (a_n b_n) = a b$.
It is a sort of definition that the real numbers are complete.
A Cauchy sequence is a sequence of real numbers $\lbrace a_n\rbrace _{n \ge 0}$ such that for any $\epsilon > 0$, there exists a sufficiently large $N_\epsilon > 0$ such that
$$ \lvert a_n - a_m \rvert < \epsilon $$
for all $n, m \ge N_\epsilon$.
Every Cauchy sequence in $\mathbb{R}$ has a limit.
Let $a_0, a_1, \ldots$ be a bounded nondecreasing sequence of real numbers. In other words, $a_i \le a_{i+1}$ for all $i \ge 0$, and there exists a universal constant $C$ such that $a_i \le C$ for all $i \ge 0$. Show that the limit $\lim_{n \to \infty} a_n$ exists.
Let $S \subseteq \mathbb{R}$ be a nonempty subset such that there exists a constant $C$ such that $x < C$ for all $x \in S$. Show that there exists a real number $M$ such that
- (i)
$x \le M$ for all $x \in S$,
- (ii)
for any $M^\prime < M$, there exists an $x \in S$ such that $x > M^\prime$.
Such $M$ is called the supremum of $S$ and we write $M = \sup S$.
We have thus defined the limit of a sequence of real numbers. Let us now define a limit of vectors.
Let $\lbrace v_k\rbrace _{k \ge 0}$ be a sequence of vectors in $\mathbb{R}^n$. Write $v_k = (v_{k,1}, \ldots, v_{k,n}) \in \mathbb{R}^n$. We say that the limit of the sequence is $x = (x_1, \ldots, x_n) \in \mathbb{R}^n$ if
$$ x_i = \lim_{k \to \infty} v_{k,i} $$
and all the limits exist. In this case, we write
$$ x = \lim_{k \to \infty} v_k. $$
Consider $\lbrace z_k\rbrace _{k \ge 0}$ be a sequence of complex numbers. Show that the following are equivalent:
- (1)
When $\mathbb{C}$ is considered as a real vector space, $\lim_{k \to \infty} z_k = z$.
- (2)
For every $\epsilon > 0$, there exists a sufficiently large $N_\epsilon > 0$ such that $\lvert z_n - z \rvert < \epsilon$ for all $n > N_\epsilon$.
But what about in general vector spaces that are not canonically isomorphic to $\mathbb{R}^n$?
Let $T : \mathbb{R}^n \to \mathbb{R}^m$ be an $\mathbb{R}$-linear map. If $\lbrace v_k\rbrace _{k \ge 0}$ is a sequence with $x = \lim_{k \to \infty} v_k$, then
$$ Tx = \lim_{k \to \infty} Tv_k. $$
Write $T$ as a matrix $T = (t_{ij})$, and let $v_k = (v_{k,1}, \ldots, v_{k,n})$. Then the $i$th component of $T v_k$ is given by
$$ (T v_k)_i = \sum_{j}^{} t_{ij} v_{k,j}. $$
Because $\lim_{k \to \infty} v_{k,j} = x_j$, we have
$$ \lim_{k \to \infty} (T v_k)_i = \sum_{j}^{} t_{ij} \lim_{k \to \infty} v_{k,j} = \sum_{j}^{} t_{ij} x_j = (Tx)_i $$
by the commutation of limits with addition and multiplication. This shows that $Tx = \lim_{k \to \infty} T v_k$.
Let $V$ be a finite-dimensional vector space over $\mathbb{R}$. For a sequence $\lbrace v_k\rbrace _{k \ge 0}$ of vectors in $V$, we say that
$$ v = \lim_{k \to \infty} v_k $$
if $Tv = \lim_{k \to \infty} Tv_k$ for a choice of isomorphism $T : V \to \mathbb{R}^n$.
What if we choose another isomorphism $S : V \to \mathbb{R}^n$? Because $S T^{-1} : \mathbb{R}^n \to \mathbb{R}^n$ is a linear map, applying $ST^{-1}$ shows that $Tv = \lim_{k \to \infty} Tv_k$ implies $Sv = \lim_{k \to \infty} Sv_k$. Also, applying $TS^{-1}$ implies the other direction. Therefore $Tv = \lim_{k \to \infty} Tv_k$ if and only if $Sv = \lim_{k \to \infty} Sv_k$. This means that the limit does not depend on the choice of isomorphism $T : V \to \mathbb{R}^n$.
Let $V$ and $W$ be finite-dimensional vector spaces over $\mathbb{R}$. We say that a function $f : V \to W$ (not necessarily a linear map) is continuous if $v = \lim_{k \to \infty} v_k$ implies $f(v) = \lim_{k \to \infty} f(v_k)$.
What we proved in Proposition 5.1.5 is that a linear map is always continuous.
Let $V, W, U$ be finite-dimensional vector spaces over $\mathbb{R}$. Let $f : V \to W$ be a continuous map, and let $g : W \to U$ be a continuous map. Show that $g \circ f : V \to U$ is continuous as well.
Show that the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \sqrt{\lvert x \rvert}$ is continuous.
We end with a useful theorem.
Let $V$ be a finite-dimensional vector space over $\mathbb{R}$, and let $S \subseteq V$ be a nonempty subset satisfying the following two properties:
- (a)
The set $S$ is bounded, i.e., there exists an isomorphism $T : V \to \mathbb{R}^n$ and a large real number $L$ such that $S \subseteq T^{-1}([-L, L]^n)$.
- (b)
The set $S$ is closed, i.e., for every sequence $\lbrace v_k\rbrace _{k \ge 0}$ with a limit, if $v_k \in S$ then $\lim_{k \to \infty} v_k \in S$.
Then for any continuous function $V \to \mathbb{R}$, the function has a maximal value on $S$. That is, there exists an $x \in S$ such that $f(x) \ge f(s)$ for all $s \in S$.