# If a function follows another one's range order, can we say it follows some continuity properties?

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.

No, let $$X = \{0\} \cup \{\frac{1}{n}: n \in \mathbb{N}\}$$, define $$f(t) = -t$$ for all $$t \in X$$, and define $$g(t) = \begin{cases}0&t > 0\cr 1&t = 0\end{cases}$$.