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.