Spivak Thomae's Function proof explanation

[fresh_42]

Sorry( for nagging), but how can we prove? Thought about it but I feel that I can't come up with my own proof.

(1) ⇒ (2) is obvious.
In the other direction simply consider ε and ε' = ε - 1/n for sufficiently large n.

 

[Samy_A]

Sorry( for nagging), but how can we prove? Thought about it but I feel that I can't come up with my own proof.

Let's prove that
(1) $\forall \epsilon>0 \ \exists \delta>0: \forall x: |x-a|<\delta \ \Rightarrow |f(x)-f(a)|<\epsilon$
implies
(2) $\forall \epsilon>0 \ \exists \delta>0: \forall x: |x-a|\leq \delta \ \Rightarrow |f(x)-f(a)|\leq \epsilon$.

Take $\epsilon>0$
From (1) we know that $\exists \delta_1>0: \forall x: |x-a|<\delta_1 \ \Rightarrow |f(x)-f(a)|<\epsilon$
Set $\delta = \delta_1/2$
Then if $|x-a|\leq \delta$, surely $|x-a|<2*\delta=\delta_1$, so that $|f(x)-f(a)|<\epsilon$
But if $|f(x)-f(a)|<\epsilon$, surely $|f(x)-f(a)|\leq\epsilon$
Hence we have deduced (2) from (1)

EDIT: @fresh_42 has given you a hint for (2) implies (1).

 

[fresh_42]

The idea is simple, therefore I called it obvious (in both directions). In every ball (interval) with positive radius (length) you can always find another ball (interval) within with a smaller but still positive radius (length). Therefore it doesn't matter whether you consider the balls (intervals) with or without their surface (boundaries). In general topological spaces, however, continuity is defined without the surfaces / boundaries because they are not always as convenient as $ℝ$ or $ℝ^n$. This has been the reason for Samy's and mine short debate on the issue. And it is the reason why Samy's Analysis professor insisted on "<". To his excuse I'd like to mention that it's been in the 1st year so he probably was especially fussy.

 

[Samy_A]

The idea is simple, therefore I called it obvious (in both directions). In every ball (interval) with positive radius (length) you can always find another ball (interval) within with a smaller but still positive radius (length). Therefore it doesn't matter whether you consider the balls (intervals) with or without their surface (boundaries). In general topological spaces, however, continuity is defined without the surfaces / boundaries because they are not always as convenient as $ℝ$ or $ℝ^n$. This has been the reason for Samy's and mine short debate on the issue. And it is the reason why Samy's Analysis professor insisted on "<". To his excuse I'd like to mention that it's been in the 1st year so he probably was especially fussy.

I agree with everything, except the last part: my professor was always fussy.
But I came to like him and appreciate him and his teaching. From day one he made it clear to our poor student brains that Mathematics has to be rigorous, and it was a lesson well learned. I may cut an edge sometimes now, but only when I know it is Ok to do so.

 

[Alpharup]

But if $|f(x)-f(a)|<\epsilon$, surely $|f(x)-f(a)|\leq\epsilon$
Hence we have deduced (2) from (1)

EDIT: @fresh_42 has given you a hint for (2) implies (1).

I can't understand this step

 

[Samy_A]

I can't understand this step

$A<\epsilon$ means that $A$ is smaller than $\epsilon$.
$A\leq \epsilon$ means that $A$ is smaller than or equal to $\epsilon$.
Any $A$ that satisfies $A< \epsilon$ will also satisfy $A\leq \epsilon$.

That's all that there is to that step.