Is Trichotomy Necessary to Prove Limit Inequalities in Metric Spaces?

[tronter]

Let X be a metric space, let a \in X be a limit point of X, and let f: X \to \mathbb{R} be a function. Assume that the limit of f exists at a. Fix t \in \mathbb{R}. Suppose there exists r > 0 such that f(x) \geq t for every x \in B_{r}(a) \backslash \{a \}; then \lim_{x \to a} f(x) \geq t.

How would you prove this? Would you use Trichotomy?

 

[Office_Shredder]

Find a sequence x_k such that x_k converges to a. Then what can you say about f(x_k) for each k?

 

[boombaby]

you will get a contradiction pretty easy if lim f(x)<t.