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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 \(\displaystyle m \gt n\),

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

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

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):

\(\displaystyle | x_1 + x_2 + \ ... \ ... \ + x_n | \le |x_1| + |x_2| + \ ... \ ... \ + |x_n|\)

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

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

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

Hope that someone can help,

Peter

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 \(\displaystyle m \gt n\),

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

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

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):

\(\displaystyle | x_1 + x_2 + \ ... \ ... \ + x_n | \le |x_1| + |x_2| + \ ... \ ... \ + |x_n|\)

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

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

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

Hope that someone can help,

Peter

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