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Hi I was reading Rudin's Princiaples of MathematicalA Analysis. In the third chapter in tying to prove

If $\{p_n\}$ is a sequence in a metric space $X$ and if $\{p_n\}$ converges $\implies$ $\{p_n\}$ is bounded

The proof goes as follows Suppose $p_n \rightarrow p$. There is an integer $N$ such that $\forall n>N \implies d(p_n,p)<1$

Now if we put $r=max\{1,d(p_1,p),d(p_2,p),.....,d(p_N,p)\}$

This implies $d(p_n,p)\leq r \forall n\in N $

Why can't say $d(p_1,p)$ be infinite which makes r infinite and therefore $\{p_n\}$ unbounded?

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    $\begingroup$ In a metric space the distance $d$ is a mapping $d \colon X \times X \to \mathbb{R}$ and infinity is not a real number, so there are no "infinite" distances. $\endgroup$
    – J. J.
    Sep 3, 2014 at 7:42
  • $\begingroup$ Yes ofcourse , my bad! Thank you J.J $\endgroup$
    – user3503589
    Sep 3, 2014 at 11:30

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The definition of a metric space requires that the distances are real numbers. Infinity is not a real number, so $d(p_1,p)$ cannot be $\infty$.

There are generalizations of metric spaces where we allow infinite distances, but that's a different story.

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