## Overview This lecture connects sequences to the sets that contain them. We first prove that every convergent sequence is bounded, then develop a powerful sequential characterization of dense sets: a set is dense in $\mathbb{R}$ exactly when every real number is the limit of some sequence drawn from the set. We establish an order-preservation result — a convergent sequence of non-negative terms has a non-negative limit — and use it to show that closed intervals contain the limits of their sequences. This motivates the general definition of a **closed set** as one that contains all of its sequential limits, and we test that definition against several examples. These ideas are foundational for the topology of $\mathbb{R}$ studied later. ## Key Concepts **Bounded sequence.** A sequence $(a_n)$ is *bounded* if there exists a real number $M$ such that $$|a_n| \le M \quad \text{for all } n,$$ equivalently $-M \le a_n \le M$ for all $n$. The value $M$ "sandwiches" the entire sequence. **Convergent $\Rightarrow$ bounded.** If $(a_n)$ converges, then $(a_n)$ is bounded. The proof splits the sequence into a tail (controlled by convergence) and a finite head (controlled by taking a maximum): - From convergence with $\varepsilon = 1$, there is an index $N$ with $|a_n - a| < 1$ for all $n \ge N$. By the triangle inequality, $|a_n| = |(a_n - a) + a| \le |a_n - a| + |a| < 1 + |a|$ for $n \ge N$. - Only finitely many terms $a_1, \dots, a_{N-1}$ come before $N$. Set $$M = \max\{|a_1|, |a_2|, \dots, |a_{N-1}|, \, 1 + |a|\}.$$ A finite list always has a maximum (an infinite set need not — e.g. $(0,1)$ has no largest element), so $M$ is well defined and bounds every term. **A set containing a sequence.** A set $S$ *contains* the sequence $(a_n)$ if $a_n \in S$ for every $n$ — equivalently, the range (image) of the sequence is a subset of $S$. A sequence is a function on $\mathbb{N}$, not itself a set, so "contains" refers to its range. **Dense set — sequential characterization.** Recall $S$ is *dense* in $\mathbb{R}$ if every open interval $(a,b)$, however small, contains a point of $S$. Equivalently: $$S \text{ is dense in } \mathbb{R} \iff \text{for every } x \in \mathbb{R} \text{ there is a sequence } (a_n) \text{ in } S \text{ with } \lim_{n\to\infty} a_n = x.$$ Example: $\mathbb{Q}$ is dense in $\mathbb{R}$, so any real number — even an irrational like $\sqrt{2}$ — is the limit of a sequence of rationals. **Order limit theorem (non-negativity).** If $a_n \ge 0$ for all $n$ and $\lim a_n = a$, then $a \ge 0$. The limit cannot "fall below" a bound that every term respects. **Closed interval contains its limits.** If $(c_n)$ is a sequence in the closed interval $[a,b]$ and $c_n \to c$, then $c \in [a,b]$. This fails for open intervals: $\left(\tfrac1n\right)$ lies in $(0,2)$ but converges to $0 \notin (0,2)$. **Closed set (definition).** A set $S$ is *closed* if, for every sequence $(a_n)$ in $S$ that converges, the limit also lies in $S$. The result above says every closed interval is a closed set (under the standard topology on $\mathbb{R}$). - $\mathbb{Q}$ is **not** closed: a sequence of rationals can converge to $\sqrt{2} \notin \mathbb{Q}$. - $(0,1]$ is **not** closed: $\left(\tfrac1n\right)$ stays inside but converges to $0 \notin (0,1]$. - $(-\infty, 1]$ **is** closed; so are $\mathbb{R}$ and $\varnothing$ (vacuously — it contains no sequences at all). ## Worked Example **Claim.** If $a_n \ge 0$ for all $n$ and $\lim a_n = a$, then $a \ge 0$. **Proof (by contradiction).** Suppose instead $a < 0$. Then $-\tfrac{a}{2} > 0$, so we may take it as our $\varepsilon$. By convergence there is an index $N$ with $|a_n - a| < -\tfrac{a}{2}$ for all $n \ge N$, which unfolds and rearranges as: $$ \begin{aligned} \left|a_n - a\right| &< -\tfrac{a}{2} \\ \tfrac{a}{2} &< a_n - a < -\tfrac{a}{2} \\ \tfrac{3a}{2} &< a_n < \tfrac{a}{2} < 0. \end{aligned} $$ Thus $a_n < 0$ for all $n \ge N$ — contradicting $a_n \ge 0$. Hence $a \ge 0$. $\blacksquare$ **Application.** To show $[a,b]$ is closed, take $(c_n)$ in $[a,b]$ with $c_n \to c$. Then $b - c_n \ge 0$ and $c_n - a \ge 0$ for all $n$, so by the theorem $b - c = \lim(b - c_n) \ge 0$ and $c - a = \lim(c_n - a) \ge 0$. Therefore $a \le c \le b$, i.e. $c \in [a,b]$. ## Summary - A sequence is bounded iff some $M$ satisfies $|a_n| \le M$ for all $n$; **every convergent sequence is bounded** (control the tail with $\varepsilon=1$, the finite head with a max). - A set contains a sequence when the sequence's range is a subset of the set. - $S$ is **dense** in $\mathbb{R}$ iff every real number is a limit of a sequence from $S$; in particular every real is a limit of rationals. - **Order limit theorem:** a convergent sequence of non-negative terms has a non-negative limit; proved by assuming a negative limit and trapping a term below zero. - A set is **closed** when it contains the limit of every convergent sequence it holds. Closed intervals (and $\mathbb{R}$, $\varnothing$) are closed; $\mathbb{Q}$ and half-open intervals like $(0,1]$ are not.

Lecture 7 (Real Analysis 2020) Part 1

Quiz

✦ Hall of Aces

Discussion