Suppose that $(X,d)$ be a complete metric space, $f:X \to \mathbb{R}$ is a sequentially lower monotone function. Let $g: X \to \mathbb{R}$ be a function with the property: $f(x)\leq f(y) \Rightarrow g(x)\leq g(y), \forall x,y \in X.$
Can we say $g$ is also sequentially lower monotone?


**Remark:** A function $h:X \to \mathbb{R}$ is said to be sequentially lower monotone on a point $x\_0 \in X$, if for all sequence $\{ x\_n \}\_{n \in \mathbb{N}}$ with $h(x\_{n+1})\leq h(x\_n)$ (for large n's), which converge to $x\_0$, we have $h(x\_0)\leq\liminf\_{n \to \infty}h(x\_n)$. A function is said to be sequentially lower monotone, if it is sequentially lower monotone in each $x \in X$. It is an extension of lower semi-continuity. The answer of the question is negative for lower semi-continuous functions.



If a function follows another one's range order, can we say it follows some continuity properties?
No, let $X = \{0\} \cup \{\frac{1}{n}: n \in \mathbb{N}\}$, define $f(t) = -t$ for all $t \in X$, and define $g(t) = \begin{cases}0&t > 0\cr 1&t = 0\end{cases}$.


