Question on equivalent characterization of locally constant function A function is defined to be locally constant iff for a point there exists a neighborhood such that the function is constant on that neighborhood.
Is this equivalent to the following: Given any (convergent) sequence $x_n \to x$ then $f(x_n)$ is eventually constant?
It is clear that if $f$ is locally constant then $f(x_n)$ is eventually constant. What's not clear to me is if the other direction also holds. 
 A: The converse only holds if the domain of the map has nice properties (being first countable is sufficient, but there may be spaces where the converse holds without them being first countable).
The direction that local constantness implies eventual constantness of $f(x_n)$ for every convergent sequence $(x_n)$ is clear: Let $x = \lim x_n$. By assumption, there is a neighbourhood $U$ of $x$ on which $f$ is constant, and by the definition of convergence, we have $x_n \in U$ for all $n \geqslant n_0$, hence $f(x_n) = f(x_m)$ for all $n,m \geqslant n_0$.
We prove that the other direction holds in first countable spaces by proving the contrapositive. So let $X$ a first countable space, and $f\colon X \to Y$ a map that is not locally constant. Then there is an $x\in X$ such that $f$ is not constant on any neighbourhood $U$ of $x$. Since $X$ is first countable, there is a countable neighbourhood basis $\{ U_n : n \in \mathbb{N}\}$ at $x$, and we may assume $U_{n+1} \subset U_n$ for all $n$ (else consider $V_n = \bigcap\limits_{k\leqslant n} U_k$). In a metric space, we could choose e.g. open balls of shrinking radius: $U_n = B_{1/n}(x)$.
Then we construct a sequence converging to $x$ such that $f(x_n)$ is not eventually constant: For $n\in\mathbb{N}$, there are points [at least one] $y \in U_n$ with $f(y) \neq f(x)$. Let $x_{2n}$ be such a point, and let $x_{2n+1} = x$. For any neighbourhood $V$ of $x$, there is an $m\in\mathbb{N}$ with $U_m \subset V$. For $k \geqslant 2m$, we have $x_k \in U_m \subset V$, so $x_k \to x$. Further, by construction we have $f(x_{2k}) \neq f(x_{2k+1})$ for all $k$, so $\bigl(f(x_k)\bigr)$ is not eventually constant.

If the space is not first countable, the other direction does not in general hold when considering only convergent sequences $x_n \to x$, but it would hold if one considers convergent nets or filters instead. The proof for filters is direct from the definitions, since the neighbourhood filter of $x$ is the coarsest filter converging to $x$, and the proof for nets is similar to the proof with sequences for the first countable space.
