Question

Help pls!!
Set \(\overline{N}:=N\cup\{\infty\}\), where \(\infty\) is a "point" not in \(N\). Define\(d\):
\( \overline{N}x\overline{N}\rightarrow \mathbb{R}\) by\\\\
\(d(m,n):= |\frac{1}{n} - \frac{1}{m} |\)
\(d(n,\infty) = d(\infty, n):= \frac{1}{n} \)
\(d(\infty,\infty) := 0 \)
Show that
a) \(d\) defines a metric on\(\overline{N}\)

Answer 1

Let's see then. To be a metric, we have a few conditions we need to satisfy. First off, is \(d\) always non-negative?

Answer 2

hmm i am not sure its not given, or should we understand it from question context?

Answer 3

Well, the absolute value of anything is always non-negative, and since both 0 and \(1/n\) are non-negative, we know that \(d\) is non-negative.

Answer 4

ok... very good explanation

Answer 5

Next, we need to show \(d(m, n)=0\) iff \(m=n\). First, if \(m=n\) it's either \[|\frac{1}{m}-\frac{1}{m}|=|0|=0\]or \(d(\infty,\infty)=0\).

Answer 6

ok..

Answer 7

If \(m\neq n\), we just have to concern ourselves with \[|\frac{1}{n}-\frac{1}{m}|=|\frac{m-n}{nm}|\]However, since \(n\neq m\), this is non-zero. Hence, \(d(m, n)=0\) iff \(m=n\).

Answer 8

Third, we need to show that \(d(m, n)=d(n ,m)\). If either \(m ,n\) are infinity, this is given to us in the definition of \(d\). However, note that \[|\frac{1}{n}-\frac{1}{m}|=|-\left(\frac{1}{m}-\frac{1}{n}\right)|=|\frac{1}{m}-\frac{1}{n}|\]So we've proven that part.

Answer 9

hm is it enough for this question or do we need to show more things ?

Answer 10

Finally, we need to show that \(d(m, o)\leq d(m, n)+d(n ,o)\) for all \(m ,n, o\in\mathbb{N}\)

Answer 11

ok..

Answer 12

Fortunately, we know that \[|x+y|\leq|x|+|y|\]for all real numbers, so this means that \[\left|\frac{1}{o}-\frac{1}{m}\right|\leq\left|\frac{1}{n}-\frac{1}{m}\right|+\left|\frac{1}{o}-\frac{1}{n}\right|\]Since \[\frac{1}{n}-\frac{1}{m}+\frac{1}{o}-\frac{1}{n}=\frac{1}{o}-\frac{1}{m}\]

Answer 13

Now you also need to include a case about infinities, where one or two of the numbers is infinity.

Answer 14

hmm..

Answer 15

Or all three are infinity. If this is the case, it's easy to see that it's all 0, so no problem there.

Answer 16

If \(o=\infty\), then we have \[d(m,\infty)=\frac{1}{m}\]\[d(m, n)+d(n ,o)=\left|\frac{1}{m}-\frac{1}{n}\right|+\frac{1}{n}\]which is fairly obviously greater than or equal to \(1/m\).

Answer 17

If \(n=o=\infty\), we get that \[d(m, \infty)=\frac{1}{m}\]and \[d(m, \infty)+d(\infty, \infty)=\frac{1}{m}+0\]And \[\frac{1}{m}\leq\frac{1}{m}\]

Answer 18

Therefore, \(d\) is a metric.

Answer 19

ok cool... you have explained it very good, i will study it again next time..

Answer 20

i have also part b) for same question, i will post it right now, maybe its wont take too much your time as this question...