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I need to prove the following

Let{x$_n$} and {y$_n$} be two sequences converging in a metric space to x and y respectively. Show that d(x$_n$, y$_n$) converges to $d(x,y) $

The textbook answer has gone about putting $a_n=d(x_n,y_n)$ and then showing that

$|d(a_n,a_m)|\leq d(x_n,x_m)+d(y_n,y_m)$ ...and then proceeded for the proof

Request guide if a shortened version as given below is correct:

To prove: $|d(x_n,y_n) - d(x,y)| \leq \epsilon $ for all $n\geq N $

$d(x_n,y_n)\leq d(x_n,x) + d(x,y) + d(y,y_n)$ (by triangle inequality)(1)

Since {x$_n$} is a convergent sequence converging to $x $

$d(x_n, x)\leq \frac{\epsilon}{2}$ for all $n \geq N_1$ similarly

$d(y_n, y)\leq \frac{\epsilon}{2}$ for all $n \geq N_2$

Choose $N$ as greater of $N_1$ and $N_2$ and putting this in (1)

$d(x_n,y_n) - d(x,y) \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $ for all $n\geq N$ (2)

In 1, interchange $x_n,y_n$ with x, y to get

$d(x,y) - d(x_n-y_n) \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $ for all $n \geq N$ (3)

2 and 3 gives

$|d(x_n,y_n) - d(x,y)| \leq \epsilon$ for all $n\geq N$. hence proved

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  • $\begingroup$ In (3), do you mean $d(x_n,y_n)$ instead of $d(x_n - y_n)$, which you wrote? $\endgroup$ – Viktor Glombik May 10 '18 at 20:40
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Your proof is fine as-is and is as short as it can be. Cannot think of a shorter way to prove that.

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Your proof is okay. Here is another way: By applying the triangle inequality twice we can show $$|d(a,b)-d(x,y)| \leq d(a,x) + d(b,y) \qquad \forall a,b,x,y.$$ To show $d(x_n,y_n) \to d(x,y)$ we can obviously show $|d(x_n,y_n)-d(x,y)| \to 0$. Applying the inequality above we get $$|d(x_n,y_n)-d(x,y)| \leq d(x_n,x) + d(y_n,y)$$ where the right hand side tends to $0$ for $n\to +\infty$. This proofs the continuity of $d(\cdot, \cdot)$.

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  • $\begingroup$ I see that $|d(x_n,y_n) - d(x,y)| = ||d(x_n,y_n)| - |d(x,y)|| \le d(x_n,y_n) - d(x,y) \le d(x_n,x) + d(x,y_n) - d(x,y_n) - d(y,y_n) = d(x_n) - d(y_n)$, but how do you get $d(x_n) + d(y_n)$? $\endgroup$ – Viktor Glombik May 10 '18 at 20:45
  • $\begingroup$ What is $d(x_n)$? $d$ is a metric, that is, $d$ is a function with two arguments! $\endgroup$ – Niklas May 11 '18 at 10:42
  • $\begingroup$ The last line is supposed to read $=d(x_n,x) - d(y_n,y)$. And my question ist how you arrive at $d(x_n,x) + d(y_n,y)$. $\endgroup$ – Viktor Glombik May 11 '18 at 14:47

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