Class on 13-10-2026

Agenda today:

• Review of limits of sequences; basic ideas and properties (30')

• Exercises 1a, 1c, discussion of exercise 5 (30')

• Independent work (15')

Last class, many people told me that building proofs is a challenging part of this course. Exercise 5 is another opportunity to build a proof.

Review of Limits

Definition: Let $( x_n )_{n \in \N}$ be a sequence of real numbers. We say that $$\lim_{n \to \infty} x_n = y$$ when, for all $\varepsilon > 0,$ there exists an $N,$ such that for all $n \ge N,$ $\lvert x_n - y \rvert \le \varepsilon.$

Another important idea: a subsequence must converge to the same limit. For instance, I mentioned in class that we can prove the following.

Proposition: Let $(x_n)$ be a sequence, and let $\alpha \colon \N \to \N$ be strictly increasing. If $(x_n)$ converges, then $$\lim_{ n \to \infty } x_{\alpha(n)} = \lim_{ n \to \infty } x_n.$$

We could prove this as follows.

Proof: Let $y$ denote the limit of $(x_n).$ I claim that the sequence $(x_{\alpha(n)})$ converges to $y.$ This means that, for any $\epsilon,$ there exists some $n_\epsilon$ such that $n \ge n_\epsilon$ implies $\lvert x_{\alpha(n)} - y \rvert < \varepsilon.$ But indeed, let $n_\varepsilon$ be such that $n \ge n_\epsilon$ implies $\lvert x_n - y \rvert < \epsilon.$ This is guaranteed because $(x_n)$ converges to $y$. Because $\alpha$ is strictly increasing, we must have $\alpha(n) \ge n$ for all $n.$ In particular, if $n \ge n_\epsilon,$ then $\alpha(n) \ge n \ge n_\epsilon.$ We conclude that $\lvert x_{\alpha(n)} - y \rvert < \epsilon,$ as desired. □

You may also prove that the same holds if $\alpha$ diverges to infinity, in the sense that for every $n \in \N,$ there exists some $m \in \N$ such that for all $k \ge m,$ $\alpha(k) \ge n.$

Exercises 1a, 1c

We walked slowly through exercises 1a and 1c in class. I write only the solution of 1c.

Exercise 5

Finally, we worked on proving the following fact.

Proposition: Suppose that $(u_n)$ converges to $u$ and $(v_n)$ converges to $v.$ Then $(u_n + v_n)$ converges to $u + v.$

At the end of the class I wrote a "textbook-style" proof that looked something like the following.

Proof: Let $\epsilon > 0.$ We will find an $n_\epsilon$ such that $$n \ge n_\epsilon \implies \lvert u_n + v_n - (u +v) \rvert < \epsilon.$$ Since $u_n$ converges to $u,$ there exists an $n_u$ such that $$n \ge n_u \implies \lvert u_n - u \rvert < \epsilon / 2.$$ Similarly, since $v_n$ converges to $v$, there exists an $n_v$ such that $$n \ge n_v \implies \lvert v_n - v \rvert < \epsilon / 2.$$ Put $n_\epsilon = \max \, \{n_u, n_v \}.$ Then, if $n \ge n_\epsilon,$ in particular $n \ge n_u$ and $n \ge n_v.$ Applying the two guarantees above gives $$\begin{align*} \lvert u_n + v_n - (u + v) \rvert & = \lvert (u_n - u) + (v_n - v) \rvert \\ & \le \lvert u_n - u \rvert + \lvert v_n - v \rvert \\ & < \epsilon / 2 + \epsilon / 2 = \epsilon, \end{align*}$$ as desired. □

Don't lose sight of the basic idea: if I can guarantee that $u_n$ will get close to $u$ and $v_n$ will get close to $v,$ then $u_n + v_n$ will get close to $u + v$.

I also suggested the problem of proving that $u_n v_n$ converges to $u v.$