Need help with simple proof, metric space, open covering.

[bobby2k]

Please take a look at the proof I added, there are some things I do not understand with this proof.

1. Does it really show that |f(x)-f(y)|≤d(x,y) for all x and y? Or does it only show that if there is an ball with radius r around x, and this ball is contained in an O in the open covering, and if y is contained in that ball. Then |f(x)-f(y)|≤d(x,y)?

2. Also please take a look at the sentence I underlined with a red line. Why can he say that
f(y)≥f(x)-d(x,y)? I could understand if he wrote r instead of f(x), that is f(y)≥r-d(x,y), why is he allowed to use r instead of f(x)?

 

[jbunniii]

In the second paragraph, there is an implicit assumption that $d(x,y) < r$. This assumption is fine since continuity at $x$ depends only on what happens near $x$, but this should have been stated explicitly, because as written, the rest is nonsense if $d(x,y) \geq r$.

Regarding the red underlined inequality: I agree with you that the immediate implication is $f(y) \geq r - d(x,y)$. This inequality is valid for any $r$ in the set $S = \{r \in \mathbb{R} | r < 1 \text{ and } d(x,y) < r \text{ and } B(x;r) \subseteq O\}$. Therefore it is also valid for $r = \sup S$, but $\sup S$ is simply $f(x)$, by definition.

 

[bobby2k]

In the second paragraph, there is an implicit assumption that $d(x,y) < r$. This assumption is fine since continuity at $x$ depends only on what happens near $x$, but this should have been stated explicitly, because as written, the rest is nonsense if $d(x,y) \geq r$.

Regarding the red underlined inequality: I agree with you that the immediate implication is $f(y) \geq r - d(x,y)$. This inequality is valid for any $r$ in the set $S = \{r \in \mathbb{R} | r < 1 \text{ and } d(x,y) < r \text{ and } B(x;r) \subseteq O\}$. Therefore it is also valid for $r = \sup S$, but $\sup S$ is simply $f(x)$, by definition.

Thank you very much! But what is it that makes the inequality still hold? What has me puzzled is that r ≤ sup S for any R in S? So it seems to me that we can't change sup S with r, since the inequality in
f(y) ≥ r - d(x,y) goes the other way?

I mean, if we have a change a part of the smaller size of the inequality with a bigger one, the inequality might not still hold?
Lets say that f(y)=7, r = 6, and d(x,y)=1, and sup S = 10
then f(y) ≥ r-d(x,y),
but we can not say that f(y) ≥ sup S - d(x,y)?

 

[jbunniii]

Thank you very much! But what is it that makes the inequality still hold? What has me puzzled is that r ≤ sup S for any R in S? So it seems to me that we can't change sup S with r, since the inequality in
f(y) ≥ r - d(x,y) goes the other way?

In general, if we have any bounded set $S$, and we are given that $x \geq y$ for every $y \in S$, then we can conclude that $x \geq \sup S$.

Proof: suppose that $x < \sup S$. Now $\sup S$ is the LEAST upper bound of $S$, so $x$ is not an upper bound of $S$. Therefore there is some $y \in S$ such that $x < y$. But this contradicts $x \geq y$ for every $y \in S$.

 

[bobby2k]

Thanks man, very elegant!

 

[jbunniii]

I mean, if we have a change a part of the smaller size of the inequality with a bigger one, the inequality might not still hold?
Lets say that f(y)=7, r = 6, and d(x,y)=1, and sup S = 10
then f(y) ≥ r-d(x,y),
but we can not say that f(y) ≥ sup S - d(x,y)?

In this example, the inequality $f(y) \geq r - d(x,y)$ is false if $r > 8$. But $\sup S = 10$. Therefore there must be some values of $r \in S$ between $8$ and $10$ for which the inequality is false. So this example is different from the situation in your proof.

 

[jbunniii]

Thanks man, very elegant!

Here's an even simpler way to put it: the condition $x \geq y$ for all $y \in S$ means exactly that $x$ is an upper bound of $S$. Therefore $x$ must be at least as big as the LEAST upper bound of $S$.

 

[bobby2k]

Here's an even simpler way to put it: the condition $x \geq y$ for all $y \in S$ means exactly that $x$ is an upper bound of $S$. Therefore $x$ must be at least as big as the LEAST upper bound of $S$.

Ah offcourse, thanks!