The proposition about metric space and limit

Question

I would like to prove this proposition, but I cannot.

$(X,d)$ : metric space

$\{a_n\},\{b_n\}$ : sequences in X

$a, b \in X$

Proposition;

If $\lim_{n \to \infty} a_n = a, \lim_{n \to \infty} b_n=b$ holds, then $\lim_{n \to \infty} d(a_n, b_n) = d(a,b)$

If I can prove $|d(a_n, b_n)-d(a,b)| \leqq d(a_n, a) + d(b_n, b)$, I will be able to prove the proposition. How can I prove this inequality?

Answer 1

Using symmetry and the standard triangle inequality, you know that

$$d (a,b) \leq d(a,a_n)+d(a_n,b_n)+d(b_n,b) $$

and

$$ d(a_n,b_n)\leq d(a_n,a)+d(a,b)+d(b,b_n). $$

These two inequalites implies your desired inequality of

$$ \vert d(a_n,b_n)-d(a,b) \vert \leq d(a_n,a)+d(b_n,b) .$$