Theorem 3.2 Rudin -- How is 0 concluded?

Question

• I'm having trouble understanding why the arbitrariness of $\epsilon$ allows us to conclude that $d(p,p')<0$. It seems we could likely conclude a value such as $\frac {\epsilon}{100}$ couldn't we?? The other idea that would normally work is the limit (as $n$ approaches $\infty$, $p$ approaches $p'$) but that would mean we are further in another sequence which would have same problem Thanks in advance

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Definition of Convergence

A sequence $\{ p_n \}$ in a metric space $X$ is said to converge if there us a point $p \in X$ with the following property: For every $ \epsilon>0$ there is an integer $N$ such that $n \ge N$ implies distance function $d(p_n, p) < \epsilon.$

Answer 1

The crucial thing to note here is that $p, p'$ are fixed points in $X$ (ie they don't depend on $n$) and hence $d(p, p') $ is non-negative specific real number and let's denote this specific non-negative number by $A$. It should now be clear that in the inequality $A<\epsilon$ the number $A$ is fixed and $\epsilon$ is arbitrary. Since $A$ is fixed it can not be chosen to be something like $\epsilon/100$. The implication $$\forall \epsilon>0,0\leq A<\epsilon\implies A=0$$ is just a fancy way (like the legalese) to mean the following obvious/trivial fact:

The only non-negative real number less than any/every positive real number is $0$.

To convince yourself that the above is trivial understand that the statement remains true even if word "real" is replaced by "rational" and this property of rationals is an immediate consequence of the following

Given any positive rational number we can find a smaller positive rational number.

And the above follows from the even more obvious fact

Given any natural number we can find a greater one, namely it's successor.

It is a trademark of authors like Rudin who convert such trivialities into seemingly difficult results via excessive use of symbolism in the name of "unambiguous / precise writing".

If these statements do not look trivial then it time to revisit the way inequalities work in these number systems.

Answer 2

The moral is that the only $x\ge 0$ satisfying $x< \epsilon$ for all $\epsilon>0$ is $0$. For, suppose that $x\ne 0$, then $x=\delta>0$. Choose $\epsilon = \frac{\delta}{2}$. Then $x\not< \epsilon$, contrary to assumption.

Because $\epsilon$ is arbitrary, the statement is true for all $\epsilon>0$, and thus we are done.

Answer 3

$d(p,p') \ge 0$.

Case 1: $d(p,p') > 0$.

Let $\epsilon = d(p,p') > 0$. Then $d(p,p') < \epsilon = d(p,p')$. That is a contradiction.

Case 2: $d(p,p') = 0$.

Then..... $d(p,p') = 0$.....