1
$\begingroup$

Theorem. Let ${p_n}$ be a sequence in a metric space $X$ and $p, p' \in X$. If ${p_n}\to p$ and ${p_n}\to p'$, then $p = p'$.

In the proof of this theorem, baby Rudin uses that $$d(p, p') < \epsilon$$ for arbritary $\epsilon > 0$ implies $$d(p, p') = 0.$$

I can't see why this is true because, in the real line, no matter how small you choose a $\epsilon > 0$, there will always be a $r \in \mathbb{R}$ such that $0 < r < \epsilon$, so $d(p, p')$ need not be zero for any choice of small $\epsilon$.

Obs.: $d$ is the distance in $X$.

$\endgroup$
  • 1
    $\begingroup$ You're basically mixing up $(\forall \epsilon > 0) (\exists r \geq 0) (r < \epsilon \wedge r \neq 0)$, which is true, and $(\exists r \geq 0) (\forall \epsilon > 0) (r < \epsilon \wedge r\neq 0)$, which is false. $\endgroup$ – Connor Harris May 18 '17 at 21:17
  • 1
    $\begingroup$ The result in the title of your question holds if the word "real" is replaced by "rational". This is basic and trivial property of order relation in rationals which is extended to reals. $\endgroup$ – Paramanand Singh May 19 '17 at 5:57
  • 1
    $\begingroup$ You can compare it with simpler statements like "if a positive integer is smaller than all prime numbers then it is $1$". $\endgroup$ – Paramanand Singh May 19 '17 at 6:00
3
$\begingroup$

Proof by contradiction:

If $d(p,p')$ were not $0$, then there would be a real number smaller than it.

$\endgroup$
3
$\begingroup$

The point is, $d(p, p')$ has to be some constant number, say, $d_0$, determined prior to the choice of $\epsilon$. In this way you cannot change it to an arbitrary $r$.

On the contrary, similar arguments can be used to prove that $d_0=d(p, p')=0$. If $d_0\neq 0$, there would be some $0 < r < d_0$. Letting $\epsilon = r$ yields a contradiction that $\epsilon < d_0 < \epsilon$.

$\endgroup$
1
$\begingroup$

Assume that $$(\forall \epsilon >0)\;\;d (a,b)<\epsilon $$ and let us prove that $$a=b $$.

suppose $a\neq b $ and put $\epsilon'=d(a,b)>0 $ then

$$d (a,b)<\epsilon'$$ $$\implies d (a,b)<d (a,b) $$

which has nosense.

or

$$(\forall \epsilon>0)\;\;0\leq d (a,b)<\epsilon$$ $$\implies $$ $$(\forall n\in \mathbb N)\;\;\;0\leq d (a,b)<\frac {1}{n+1} $$

$$\implies $$

$$0\leq d (a,b)\leq \lim_{n\to+\infty}\frac{1}{ n+1}$$ $$\implies 0\leq d (a,b)\leq 0$$ $$\implies d (a,b)=0\implies a=b $$

$\endgroup$
1
$\begingroup$

"$d(p,p') < \epsilon$ for arbitrary $\epsilon > 0$" is another way of saying this:

$d(p,p')$ is strictly less than every positive number.

Since $d$ is a metric, then $d(x,y) \ge 0$ for all $x,y \in X$. Therefore:

$d(p,p') \ge 0$ by definition, and $d(p,p')$ is strictly less than every positive number.

There's only one number that satisfies both of those conditions.

$\endgroup$
0
$\begingroup$

Notice that $\epsilon$ is arbitrarily small, so in passage to a limit we can take this distance to be zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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