Prove that right hand limit exist for monotone function

Question

I am just having a slight issue I can't explain in my argument. First, we need a candidate for our limit. Suppose $A = \{f(x) : f(x) > c\}$. We can see that A is bounded below by f(c). As A is non-empty and bounded below set $inf A = k$. Consider $k + \epsilon$. As $k + \epsilon$ is no longer greater lower bound. Thus, there exists $x_0 > c: k \leq f(x_0) < k + \epsilon$. Set $\delta = x_0 - c$. Then $f(x) < k + \epsilon$, whenever $ 0 < x < \delta + c$. But why is $f(x) \geq k$?

Answer 1

Without loss of generality, let $f:\mathbb R\to\mathbb R$ be increasing, let $x_0\in\mathbb R$, let $c=\inf\{f(x):x>x_0\}$, and let $\varepsilon>0$. Then there is $x_1>x_0$ such that $f(x_1)<c+\varepsilon$. Setting $\delta=x_1-x_0$ yields that $f(x)<c+\varepsilon$ whenever $x_0<x<x_0+\delta$.