1
$\begingroup$

This question already has an answer here:

I know this is such a basic question and its embarrassing to ask but I just don't understand why Rudin concludes such a thing. In theorem 3.2 (a) he proves that if two sequences converge to different limits then the limits are actually the same as follow:

$$ \epsilon > 0 $$ $$ n \geq N \implies d(p_n,p) < \frac{\epsilon}{2}$$ $$ n \geq N' \implies d(p_n,p') < \frac{\epsilon}{2}$$ let $n = \max(N,N')$: $$ d(p,p') \leq d(p,p_n) + d(p_n,p') < \epsilon $$

everything so far made sense. The last step of the proof is what I don't understand:

since $\epsilon$ arbitrary, we conclude that $d(p,p') = 0$

thats the part that I don't understand. The proof started assuming that $ \epsilon > 0$ so of course the proof didn't conclude by plugging in zero. In chapter 2 there was such a big emphasis that the points in the neighborhoods to the limit points are different from the limit point. See this proof I'd conclude there is always a positive difference between p and p' but its never exactly zero (just like in chapter 2 and limit points).

Of course I assume I am wrong and there some subtle but important point that I don't understand and wanted to clarify it. Can someone clarify why my assertion that the difference is always positive (of course by assumption) is wrong? If anything I'd conclude that $p$ and $p'$ are limits of each other but not that they are the same as Rudin concludes or that their distance is zero. What did I miss?


note: I do know limit points and limits aren't the same, was just saying that to provide context for my confusion.

$\endgroup$

marked as duplicate by Paramanand Singh limits Apr 24 '18 at 3:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Because the open interval $(0,\infty)$ doesn't have a minimum element but is bounded below with infimum $0$. $\endgroup$ – Prasun Biswas Apr 23 '18 at 15:08
  • 1
    $\begingroup$ To be more explicit, if $d(p,p')=k>0$, then choosing $\epsilon$ such that $0\lt\epsilon\le k$ gives a contradiction. $\endgroup$ – Prasun Biswas Apr 23 '18 at 15:12
  • $\begingroup$ @PrasunBiswas I think a better answer is the one I already provided, by trichotomy. $\endgroup$ – Pinocchio Apr 23 '18 at 15:12
  • $\begingroup$ Pinocchio. Assume the contrary: d(p,p')=a >0, then, pick an epsilon =a/2, say. What do you get ? d(p,p') =a <a/2 (=epsilon).Contradiction.ok? $\endgroup$ – Peter Szilas Apr 23 '18 at 15:12
  • 2
    $\begingroup$ I think the simplest answer would be that $0$ is the only non-negative real number that is less than any positive real number. $\endgroup$ – Prasun Biswas Apr 23 '18 at 15:15
1
$\begingroup$

Others explained this but basically you have

$$0\leq d(p,p')<\epsilon \ \forall \epsilon>0$$

If you let $d(p,p')>0$ then just pick $\epsilon = d(p,p')/2$ to get a contradiction. Hence the only value $d(p,p')$ can take on is zero.

$\endgroup$
  • $\begingroup$ I don't think one needs contradiction: "I think the simplest answer would be that 0 is the only non-negative real number that is less than every positive real number." $\endgroup$ – Pinocchio Apr 23 '18 at 15:17
  • $\begingroup$ True, and how would you prove that fact if you needed to? Once you know it, it is fine to reference it without the above explanation, but you should see the proof for it once $\endgroup$ – pureundergrad Apr 23 '18 at 15:17
  • $\begingroup$ by trichotomy as I provided in my answer ;) [not contradiction] $\endgroup$ – Pinocchio Apr 23 '18 at 15:18
  • $\begingroup$ but you deserve an upvote, your answer is good regardless, thanks! :) $\endgroup$ – Pinocchio Apr 23 '18 at 15:20
1
$\begingroup$

Another way of putting it and this is probably more likely in Rudin's thinking:

$\mathbb R$ has the least upper bound property and the basic definition of "bounded below", "lower bound" and "greatest lower bound" make the following statement almost a tautology.

Let $S\subset \mathbb R$ be a non-empty set that is bounded below. And let $K < s$ for any $s \in S$. Then that would mean $K \le \inf S$. (Because if $K < s$ for all $s\in S$ then $K$ is a lower bound of $S$ so $K \le \inf S$, the greatest lower bound. [We do require that the least upper bound property holds on $\mathbb R$ however.])

Now let $S= \mathbb R^+ = \{x \in \mathbb R| x > 0\}$. $\mathbb R^+$ is bounded below by any non-positive number. So $\inf \mathbb R^+$ exists and it takes little effort to discover $\inf \mathbb R^+ = 0$.

$K < \epsilon$ for any $\epsilon \in \mathbb R^+$ means $K \le \inf \mathbb R^+ = 0$.

If we are further told $K \ge 0$ the we have $K \le 0$ and $K \ge 0$. That means..... $K = 0$.

.....

But that's a lot of sound and fury for what should be intuitive once you think about it:

...

Consider the statement. "Let $K$ be a non-negative real number that is smaller than any arbitrary positive number".

It's subtle but that statement is possible but only if $K = 0$.

If such a number exists it can't be negative because we were told $K$ is not negative.

It can't be positive because it is smaller than any positive number so if it were positive it would be smaller than itself, which is a logic impossibility-- no number can be smaller than itself.

The only third option is $K = 0$ and that is not inconsistent. $0$ is non-negative, and $0 < \epsilon$ for all $\epsilon > 0$.

...

$\endgroup$
  • $\begingroup$ The last part is simpler and this is my preferred way +1 $\endgroup$ – Paramanand Singh Apr 24 '18 at 3:04
  • $\begingroup$ True. But somehow it seems subtle and ... "gee, there must be some trick going on"... But if you realize "$K < \epsilon$ for all positive epsilon" $\iff$ "$K$ is a lower bound of the set of all positive numbers", well, then it's just common sense... $\endgroup$ – fleablood Apr 24 '18 at 19:27
0
$\begingroup$

I think I get it now. Since

$$ d(p,p') < \epsilon $$

(and $ 0 \leq d(p,p')$)

for all positive numbers then by trichotomy (i.e. one has to hold, <,>,=) then it must mean that its not equal to any positive or negative number then it must be zero (because we know all the things is not and since 1 thing does have to hold it must be equality to zero).

"I think the simplest answer would be that 0 is the only non-negative real number that is less than every positive real number."


On a related statement I noticed that if we change things to less than or equal that $d(p,p')=0$ is still zero.

$$ d(p,p') \leq \epsilon $$

Proof:

Assume $\forall \epsilon > 0, d(p,p') \leq \epsilon$ AND $d(p,p') \neq 0$. Then there must exist some $\delta >0$ s.t. $d(p,p') = \delta$. Then the $d(p,p') \leq \epsilon$ can't hold for every epsilon because it doesn't hold for $\frac{ \delta}{2}$. QED.

$\endgroup$
  • $\begingroup$ Your last highlighted sentence is the key and it should be obvious. The thing is made difficult only by use of Greek symbols. $\endgroup$ – Paramanand Singh Apr 23 '18 at 15:18
  • $\begingroup$ @ParamanandSingh I really must insist that trichotomy is the answer. Its a property of ordered sets in definition 1.5 (i) in Rudin. We don't need contradiction to prove it. The quote I highlighted was to make the intuitive answer highly visible but its just a summary of trichotomy. $\endgroup$ – Pinocchio Apr 23 '18 at 15:20
  • $\begingroup$ Trichotomy works when you are comparing two specific numbers. That does not apply here. Contradiction is the only way out. $\endgroup$ – Paramanand Singh Apr 23 '18 at 15:22
  • $\begingroup$ we are comparing two numbers. $d(p,p')$ vs every epsilon greater than zero and concluding that $d(p,p') < \epsilon$ holds for all of them. The same for negative numbers. But since $d(p,p')$ does have some value (greater than or equal to zero) and we've ruled out all the numbers is not what is left is what it is. No need to do contradiction. $\endgroup$ – Pinocchio Apr 23 '18 at 15:35
  • $\begingroup$ We are not comparing those quantities, rather it is given that the relation of less than holds. So trichotomy is not helping us here. The contradiction is needed to answer the question "why can't d(p, p') be positive" $\endgroup$ – Paramanand Singh Apr 23 '18 at 15:41

Not the answer you're looking for? Browse other questions tagged or ask your own question.