Limit of a function with discrete codomain

Question

Let $f(n) = \sum_{i=0}^n 2^{-i}$

Let $g(n) = \begin{cases} \text{False}, & f(n) < 2\\ \text{True}, & f(n) \geq 2 \end{cases}$

I’m trying to find $\lim_{n \to \infty} g(n)$

Clearly $\lim_{n \to \infty} f(n) =2$

I do not think it is valid to say that because $f(n)=2$ in the limit then $g(n) = \text{True}$ in the limit.

For any finite $n$ the value of $g(n)$ is False which leads me to believe that the limit is False but I’m sure about this.

Possibly this depends on the domain of $f$. If the domain of $f$ is $\mathbb{N} \cup \{\infty\}$ then is the limit different to the case when the domain of $f$ is $\mathbb{N}$?

Answer 1

You are correct: $\lim\limits_{n\to\infty}g(n)=\text{False}$. This is true whether $f$ is defined on $\Bbb N$ or on $\Bbb N\cup\{\infty\}$. Just as the value of $\lim\limits_{x\to 0}h(x)$ does not depend on the value of $h$ at $x=0$ (which may not even be defined), so the limit here as $n\to\infty$ does not depend on the value of $g$ at $\infty$. The sequence $\langle g(n):n\in\Bbb N\rangle$ is constant with value $\text{False}$, so its limit is $\text{False}$ irrespective of whether $g(\infty)$ is defined or, if so, what its value is.

Answer 2

$\forall n \in \mathbb{N}, g(n)=false$

Hence, $\lim_{n \rightarrow + \infty} g(n) = false$

If you ever have a doubt, you can go back to the definition of the limit. If you consider the norm on booleans being either $0$ or $1$

Let us consider $\epsilon > 0$

Let us call $n_0=0$

Let us consider $n \geq n_0 $

$|g(n) - false|<\epsilon$

So $\forall \epsilon > 0, \exists n_0 \in \mathbb{N}, \forall n \geq n_0, |g(n) - false | < \epsilon $

Following a comment : If the domain of $f$ and $g$ includes $\infty$, then this does not apply, and indeed $g(\infty)=true$, but $\infty$ will not have the same meaning than in the limit. It will be part of the function domain. Although it is the same idea behind both infinities, one is defined thanks to the limit, and the other as some object, on which $f$ equals its limit and $g$ equals true.

By posing $f(-1) = \lim_{n \rightarrow \infty} f(n)$ (and $g(-1) = true$) the same thing would appear, but be maybe less confusing.