Metric Spaces - Fixed Point Theorem (Apostol, Theorem 4.48)

[Math Amateur]

I need help with the proof of the Fixed Point Theorem for a metric space (S,d) (Apostol Theorem 4.48)

The Fixed Point Theorem and its proof read as follows:View attachment 3901
View attachment 3902
In the above proof Apostol writes:

" ... ... Using the triangle inequality we find for $$m \gt n$$,

$$d(p_m, p_n) \le \sum_{k=n}^{m-1} d(p_{k+1}, p_k ) $$ ... ... "I am unsure of how (exactly!) Apostol uses the triangle inequality to derive the relation

$$d(p_m, p_n) \le \sum_{k=n}^{m-1} d(p_{k+1}, p_k ) $$

Can someone please help by showing how (formally and rigorously) this is derived?

I presume that Apostol is using the generalised form of the triangle inequality which he describes as the following (see page 13):

$$| x_1 + x_2 + \ ... \ ... \ + x_n | \le |x_1| + |x_2| + \ ... \ ... \ + |x_n|$$

... ... but ... ... I cannot see how he derives a situation where:

$$d(p_m, p_n) = d(p_{n+1}, p_n) + d(p_{n+2}, p_{n+1}) + \ ... \ ... \ + d( p_m, p_{m-1} )$$

so that the triangle inequality can be applied ... ...

Hope that someone can help,

Peter

 

[Fallen Angel]

Hi Peter,

The triangle inequality is valid for any metric, not only the absolute value, this is, if $(X,d)$ is a metric space, then $d(x,z)\leq d(x,y)+d(y,z)$ for all $x,y,z\in X$.

Now he applies this as many times as needed.

$d(p_{m},p_{n})\leq d(p_{n},p_{n+1})+d(p_{n+1},p_{m})$
$d(p_{n+1},p_{m})\leq d(p_{n+1},p_{n+2})+d(p_{n+2},p_{m})$

and so he can conclude $d(p_{m},p_{n})\leq \displaystyle\sum_{k=n}^{m-1}d(p_{k+1},p_{k})$

 

[Math Amateur]

Hi Peter,

The triangle inequality is valid for any metric, not only the absolute value, this is, if $(X,d)$ is a metric space, then $d(x,z)\leq d(x,y)+d(y,z)$ for all $x,y,z\in X$.

Now he applies this as many times as needed.

$d(p_{m},p_{n})\leq d(p_{n},p_{n+1})+d(p_{n+1},p_{m})$
$d(p_{n+1},p_{m})\leq d(p_{n+1},p_{n+2})+d(p_{n+2},p_{m})$

and so he can conclude $d(p_{m},p_{n})\leq \displaystyle\sum_{k=n}^{m-1}d(p_{k+1},p_{k})$

Thanks Fallen Angel ...

Appreciate your help ... ...

Peter

 

[Euge]

I just want to add that the triangle inequality is an axiom for a metric. Sorry Peter if this sounds redundant.

 

[Math Amateur]

I just want to add that the triangle inequality is an axiom for a metric. Sorry Peter if this sounds redundant.

Hi Euge,

No ... that does not sound redundant at all ...

Your post is relevant and most helpful,

Thanks,

Peter