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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?

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    $\begingroup$ This works. Another way to see this is that $\mathbb{N}$ has the discrete topology (this is the topology it inherits when we embed it into $\mathbb{R}$) in which every subset is open. Therefore every function is continuous. $\endgroup$ Commented Jul 3, 2015 at 2:00
  • $\begingroup$ Notice that sequential continuity implies continuity only on $1st$-countable spaces. The converse is always true: continuity always implies sequential continuity. $\endgroup$ Commented Jul 3, 2015 at 2:40

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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.

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