This is a question that came up as a true false question in my textbook, and I was wondering what you thought of my reasoning. I claim that even though a graph of such a function doesn't look continuous, that by the sequential definition of continuity, that the function is continuous.
According to the text book, a function $f:D \to R$ is said to be continuous at the point $x_0$ provided that whenever $\{x_n\}$ is a sequence in $D$ that converges to $x_0$, the image sequence $\{f(x_n)\}$ converges to $f(x_0)$. The function is said to be continuous if this holds for all $x \in D$.
It would seem that by this definition, a function $f:\mathbb N \to \mathbb R$ would be continuous. Loosely speaking, a sequence of natural numbers has to eventually be constant in order to be convergent (i.e. {1,2,3,4,5,5,5,5,5,…}), since the smallest difference between natural numbers is $1$, and we can pick $0 < \epsilon <1$.
Therefore, the image sequence will eventually be constant as well, so no matter what $\epsilon$ we choose, we can always find an index $N$ such that $f(x_n) – f(x_0) = 0 < \epsilon$ for all indices $n \geq N$. Therefore we can always meet the requirement for convergence of $f(x_n) \to f(x_0)$ and thus $f$ is continuous.
What do you all think?
Best Answer
Your reasoning is correct. I'll explain the more popular way of doing things. There are many versions of the definition of continuity. Here are some of them: $(X,d_x)$, $(Y,d_y)$ metric spaces, $f:X \rightarrow Y$ is continuous at $x \in X$ if
(1) $\forall \epsilon > 0$ there exists $\delta > 0$ such that for all $y \in X$ such that $d(x,y) < \delta$ we have $d(f(x),f(y)) < \epsilon$
(2) For any sequence $(x_n)_{n=1}^\infty$ in $X$ converging to $x$ we have $\lim_{n \to \infty} f(x_n) = f(x)$
(If you don't know what a metric space is, its a set equipped with a "distance function" d. In your case $X = \mathbb{N}$ and $d$ can be picked among a few options).
From a topological perspective we also have, given $f:X \rightarrow Y$, $f$ is continuous if for every open set $\Omega \subset Y$ we have $f^{-1}(\Omega) \subset X$ is also open. The topological definition is the nicest to establish your claim. $N$ is usually endowed with the discrete topology: $d(x,y) = 1$ if $x\neq y$, $d(x,y) = 0$ if $x=y$. Choosing $\delta = 1/2$ it is easy to show every subset of $N$ is open. By the above definition any function on $\mathbb{N}$ is continuous.
A "discrete space" is usually defined as one in which every subset is open, therefore every function is continuous. Some nice are $\mathbb{Z}^n$, finitely generated abelian groups, etc.