Show $d(x,y) = d(y,x)$

Question

If

$$\mu (x,y) = \min\{n\in\mathbb{N} \ | \ x_n \not= y_n \}$$

and

$$d(x,y) = \frac{1}{\mu(x,y)}$$

How can I show that

$$d(x,y)=d(y,x)$$

For me it's pretty obvious, but I don't know how to show it mathematically.

Hint:It's the equivalent of showing that $μ(x,y)=μ(y,x)$.. — MathematicianByMistake, May 31 '15 at 14:21

score 1 · Accepted Answer · answered May 31 '15 at 14:22

Mathematically, it just comes from the fact that :

$$\{n\in\mathbb{N}|x_n\neq y_n\}=\{n\in\mathbb{N}|y_n\neq x_n\}$$

Now to understand why those two sets are equal it just come from the fact that $a\neq b$ if and only if $b\neq a$ which comes from the fact that $a=b$ if and only if $b=a$.

score 0 · Answer 2 · answered May 31 '15 at 14:21

0

$\mu(x,y)= min \lbrace n \in \mathbf{N}, x_{n} \not = y_{n}= min \lbrace n \in \mathbf{N}, y_{n} \not = x_{n} \rbrace=\mu(y,x)$. Is this enough ?

answered May 31 '15 at 14:21

mich95

8,713

Yes thank you, again. – Stranqer95 May 31 '15 at 14:31

score 0 · Answer 3 · answered May 31 '15 at 14:22

0

Note that : $$\min\{n \in \mathbb{N} | \ x_n \neq y_n \}=\mu (x,y) = \mu (y,x) = \min\{n \in \mathbb{N} | \ y_n \neq x_n \}$$ So $$d(x,y)=d(y,x)$$

answered May 31 '15 at 14:22

the8thone

4,111

Show $d(x,y) = d(y,x)$

3 Answers3