Example G.1.
Let \(\seq an\) and \(\seq bn\) be sequences of real numbers, and suppose that \(a_n \to 0\) and \(b_n \to 0\) as \(n \to \infty\text{.}\) Show, directly using the limit definition, that \(a_n + b_n \to 0\text{.}\)
Section G.1 When \(\forall\) comes before \(\exists\)
Suppose we need to write a proof of the following fact from Analysis 1, and we don’t have the algebra of limits to help us out.
I sometimes see students writing arguments in the following form:
Flawed argument.
For any \(\varepsilon_1 \gt 0\) we can find \(N_1 \in \N\) such that \(\abs{a_n} \lt \varepsilon_1\) for all \(n \ge N_1\text{.}\) Similarly, for any \(\varepsilon_2 \gt 0\) we can find \(N_2 \in \N\) such that \(\abs{b_n}
\lt \varepsilon_2\) for all \(n \ge N_2\text{.}\) Let \(N=\max\{N_1,N_2\}\text{.}\) Then for all \(n \ge N\text{,}\)
\begin{equation*}
\abs{a_n+b_n} \le \abs{a_n} + \abs{b_n} \lt \varepsilon_1 + \varepsilon_2 =
\varepsilon\text{.}
\end{equation*}
This argument doesn’t prove much of anything! To see why, let’s write out what we need to show:
For all \(\varepsilon \gt 0\text{,}\) there exists \(N \in \N\) such that \(\abs{a_n+b_n} \lt \varepsilon\) for all \(n \ge N\text{.}\)
In other words, someone hands you some \(\varepsilon \gt 0\text{,}\) and it is up to you to pick \(N\) appropriately. The above solution puts the cart before the horse, picking \(N\) according to some mysterious \(\varepsilon_1\) and \(\varepsilon_2\text{,}\) and then effectively defining \(\varepsilon=\varepsilon_1+\varepsilon_2\) in the very final line. This is not a valid way to argue, and would get very few marks in an exam for a high-level analysis unit such as this one.
Thankfully, in this case and many others, it’s relatively easy to fix. Most often, the lecture notes will do it in the following way:
Correct argument.
Fix \(\varepsilon \gt 0\text{.}\) Then we can find \(N_1 \in \N\) such that \(\abs{a_n} \lt
\varepsilon/2\) for all \(n \ge N_1\text{.}\) Similarly, we can find \(N_2 \in
\N\) such that \(\abs{b_n} \lt \varepsilon/2\) for all \(n \ge N_2\text{.}\) Let \(N=\max\{N_1,N_2\}\text{.}\) Then for all \(n \ge N\text{,}\)
\begin{equation*}
\abs{a_n+b_n} \le \abs{a_n} + \abs{b_n}
\lt \frac \varepsilon 2 + \frac \varepsilon 2 = \varepsilon\text{.}
\end{equation*}
Note how here there is no mystery about what \(\varepsilon\) is. It is some fixed number given to us at the outset, and everything we do from that point forward depends on \(\varepsilon\) in a straightforward way. Of course the annoying part is that we have to either guess that \(\varepsilon/2\) is the right thing to use, or (what is more likely) go back and tweak things after we’ve already written part of our argument (perhaps on scratch paper). For instance, we could have written the flawed argument, and then gone back and replaced \(\varepsilon_1\) and \(\varepsilon_2\) with \(\varepsilon/2\text{.}\)
If you really want to use use intermediate variables like \(\varepsilon_1\) and \(\varepsilon_2\text{,}\) then you have to be crystal clear about their contingent status. One way to do this is as follows:
An alternative argument.
Fix \(\varepsilon \gt 0\text{,}\) and let \(\varepsilon_1 \gt 0\) and \(\varepsilon_2 \gt
0\) be positive real numbers to be determined. Then we can find \(N_1 \in \N\) such that \(\abs{a_n} \lt \varepsilon_1\) for all \(n \ge
N_1\text{,}\) and \(N_2 \in \N\) such that \(\abs{b_n} \lt \varepsilon_2\) for all \(n \ge N_2\text{.}\) Let \(N=\max\{N_1,N_2\}\text{.}\) Then for all \(n \ge N\text{,}\)
\begin{equation*}
\abs{a_n+b_n} \le \abs{a_n} + \abs{b_n} \lt \varepsilon_1 + \varepsilon_2 = \varepsilon\text{,}
\end{equation*}
provided we choose \(\varepsilon_1 \gt 0\) and \(\varepsilon_2 \gt 0\) such that \(\varepsilon_1 + \varepsilon_2 = \varepsilon\text{,}\) say \(\varepsilon_1 = \varepsilon_2 = \varepsilon/2\text{.}\)
While all of this fussing over \(\varepsilon\)’s might seem pedantic, without it it is extremely easy to ‘prove’ things that are just not true; see Exercise B.1. I have seen this happen many times on exams for this for this unit, often with catastrophic results.
