# Thread: xn->a, yn->b imply d(xn, yn)->d(a,b)

1. ## xn->a, yn->b imply d(xn, yn)->d(a,b)

Let (x_n) and (y_n) be sequences in a metric space X. Suppose that x_n converges to a and y_n converges to b. Prove that d(x_n, y_n) converges to d(a, b).
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|d(x_n, y_n) - d(a,b)|
We need to relate this to the hypothesis, so I think perhaps subtracting and adding d(x_n, a) would help...but still I can't figure out how to prove the result...

Any help is appreciated!

2. Originally Posted by kingwinner
Let (x_n) and (y_n) be sequences in a metric space X. Suppose that x_n converges to a and y_n converges to b. Prove that d(x_n, y_n) converges to d(a, b).
================================

|d(x_n, y_n) - d(a,b)|
We need to relate this to the hypothesis, so I think perhaps subtracting and adding d(x_n, a) would help...but still I can't figure out how to prove the result...

Any help is appreciated!
$d(x_n,y_n)-d(x,y)\leqslant d(x_n,x)+d(y_n,y)-d(x,y)\leqslant$ $d(x_n,x)+d(y_n,y)+d(x,y)-d(x,y)=d(x_n,x)+d(y_n,y)$.

Also, $d(x,y)-d(x_n,y_n)\leqslant d(x,x_n)+d(y,x_n)-d(x_n,y_n)\leqslant$ $d(x,x_n)+d(y,y_n)+d(x_n,y_n)-d(x_n,y_n)=d(x_n,x)+d(y_n,y)$.

Combinging these two gives $\left|d(x_n,y_n)-d(x,y)\right|\leqslant d(x_n,x)+d(y_n,y)$.

Conclude.

3. Originally Posted by Drexel28
$d(x_n,y_n)-d(x,y)\leqslant d(x_n,x)+d(y_n,y)-d(x,y)\leqslant$ $d(x_n,x)+d(y_n,y)+d(x,y)-d(x,y)=d(x_n,x)+d(y_n,y)$.

Also, $d(x,y)-d(x_n,y_n)\leqslant d(x,x_n)+d(y,x_n)-d(x_n,y_n)\leqslant$ $d(x,x_n)+d(y,y_n)+d(x_n,y_n)-d(x_n,y_n)=d(x_n,x)+d(y_n,y)$.

Combinging these two gives $\left|d(x_n,y_n)-d(x,y)\right|\leqslant d(x_n,x)+d(y_n,y)$.

Conclude.
Why is the first inequality (d(xn,yn) d(xn,x)+d(yn,y) ) true?

4. Originally Posted by kingwinner
Why is the first inequality (d(xn,yn) d(xn,x)+d(yn,y) ) true?
Should be $d(x_n,x)+d(y_n,x)$ and the next step gets all the $x_n$'s with the $x$'s and the $y_n$'s with the $y$s