Convergent Sequences: Prove Lim d(x_n,y_n)=d(a,b)

[Kreizhn]

Homework Statement

Let (X,d) be a metric space with two sequences $(x_n), (y_n)$ which converge to values of a,b respectively. Show that

$$\lim_{n \to \infty} d(x_n,y_n) = d(a,b)$$

Homework Equations

$$(x_n) \rightarrow a \Leftrightarrow \forall \epsilon >0 \quad \exists n_0 \in \mathbb{N} \text{ such that } \forall n>n_0 \quad d(x_n,a)< \epsilon$$

$$d(x,z) \leq d(x,y) + d(y,z) \quad \forall x,y,z \in X$$

The Attempt at a Solution

This seems like it should be a fairly easy question, but I don't have much analysis in my background. I attempted to proceed as follows:

Since $d:X\times X \rightarrow \mathbb{R}$, it is sufficient to show that $\forall \epsilon >0 \quad \exists n\in \mathbb{N} \text{ such that } |d(x_n,y_n) - d(a,b)|< \epsilon$. So let $\epsilon >0$ and $n' = max\{ n_0, n_1 \}$ where $n_0, n_1$ are natural numbers which satisfy the individual convergence properties for $(x_n),(y_n)$. Let $n>n'$ giving

$$|d(x_n,y_n) - d(a,b) | &=& |d(x_n,y_n) + d(x_n,b) - d(x_n,b) - d(a,b)|<br /> <br /> \leq |d(y_n,b) - d(x_n,a)|<br /> <br /> < |\epsilon - \epsilon|<br /> <br /> \leq \epsilon$$

But I'm really not sure about the $|\epsilon - \epsilon| \leq \epsilon$ line.

Any thoughts would be appreciated.

 

[Dick]

Well, if 0<=e1<=e and 0<=e2<=e then |e1-e2|<=e. That's what you really mean by |e-e|<=e, right?

 

[Kreizhn]

Yes, that was my thought process.