Proof involving neighborhood of a limit point

Question

Let $f:D \to R$ with $x_0$ as an accumulation point of $D$. Prove that $f$ has a limit at $x_0$ if for each $\epsilon >0$ there is a neighborhood $Q$ of $x_0$ such that, for an $x,y \in Q \cap D$, $x \ne x_0$ and $y \ne x_0$, we have $\lvert f(x)-f(y) \rvert < \epsilon$.

Pf:Let $x_0$ be an accumulation point of $D$. It is true that $\lvert f(x)-L \rvert =\lvert f(x) -f(x_0) \rvert<\frac{\epsilon}{2}$ for all $x \in Q \cap D$ if $0<\lvert x-x_0 \rvert < \delta$

$$\lvert f(x)-f(y) \rvert = \lvert f(x)-f(x_0)+f(x_0)-f(y) \rvert \le \lvert f(x)-f(x_0) \rvert + \lvert f(x_0)-f(y) \rvert < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

Therefore $\lvert f(x) -f(y) \rvert < \epsilon$ for all $x,y \in Q \cap D$ such that $0<\lvert x-y \rvert < \delta $

I feel like I can word this proof better

Answer 1

Your answer is fine, but you $\delta$ argument could be a little stronger, and predicate your $\epsilon$ argument.

The general argument for $\epsilon, \delta$ argument should be phrased,

1. Let $\epsilon > 0$ be given
2. Then for $|x-y| < \delta$ the following is true,
3. Math statement $< \epsilon$

In addition, when you say, "It is true [stuff] for $x \in Q \cap D$ and $|x - x_0| < \delta$", you should really add something like "so, $|x-y| \leq |x-x_0| + |y-x_0| < \frac{\delta}{2} + \frac{\delta}{2} = \delta$ [...]"