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This excerpt relates to the following proof:

If $f$ is continous on $[a,b]$, then $f$ is uniformly continuous $[a,b]$.

For $\epsilon > 0$ let's say that $f$ is $\epsilon$-good on $[a,b]$ if there is some $\delta > 0$ such that, for all $y$ and $z$ in $[a,b]$,

if $|y-z|<\delta$ , then $|f(x)-f(z)|<\epsilon$

Consider any $\epsilon > 0$. Let

$A=\{x:a\leq x \leq b$ and $f$ is $\epsilon$-good on $[a,x]\}$

Then, the proof went on to show that $A$ has a least upper bound, $\alpha$ and that $\alpha=b$. Here's the part I don't understand:

... So $f$ is surely $\epsilon$-good on the interval $[\alpha - \delta_0, \alpha + \delta_0]$. On the other hand, since $\alpha$ is the least upper bound of $A$, it is also clear that $f$ is $\epsilon$-good on $[a, \alpha- \delta_0]$.

I don't understand how we can conclude that $f$ is $\epsilon$-good $[a, \alpha- \delta_0]$ just from knowing that $\alpha$ is the least upper bound of $A$. How do we know that $f$ is $\epsilon$-good for each number in the interval from $a$ up to and including $\alpha-\delta_0$?

Here's my understanding of the proof:

Suppose $a=1$ and 5 is the least upper bound of $A$. Then, just by knowing that $f$ is $\epsilon$-good on $[5-\delta_0,5+\delta_0]$ we can conclude that $f$ is $\epsilon$-good on the intervals $[a,2],[a,3]$ all the way to $[a, 5-\delta_0]$. I feel like I am missing some property of the least upper bound that is preventing me from understanding this proof.

The proof also claims that $f$ is $\epsilon$-good on $[a, b - \delta_0]$ which I also don't understand but I guess the explanation would be similar to $f$ being $\epsilon$-good on $[a,\alpha-\delta_0]$.

Thank you in advance for any help provided.

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It is better to define least upper bound in plain and simple English with minimal amount of symbolism.

A number $m$ is said to be the least upper bound of a non-empty subset $A$ of real numbers if no member of $A$ exceeds $m$, but given any number $m'$ less that $m$ it is exceeded by at least one member of $A$.

Now adding the context of the current problem to this definition, $\alpha$ is the least upper bound of set $$A = \{x \mid a \leq x \leq b \text{ and }f\text{ is }\epsilon\text{-good on }[a, x]\}$$

Consider the number $\beta = \alpha - \delta_{0}$. Since $\delta_{0} > 0$ it means that $\beta < \alpha$ and hence by the second part of the definition of least upper bound this number $\beta$ must be exceeded by a member of set $A$. Thus we have a $y\in A$ such that $y > \beta$. Since $y \in A$ it means that $f$ is $\epsilon$-good in $[a, y]$. Now $y > \beta$ and therefore it is clear that the interval $[a, \beta]$ is fully contained in the interval $[a, y]$. The $\epsilon$-goodness of $f$ on $[a, y]$ implies its $\epsilon$-goodness on any subinterval of $[a, y]$ and thus $f$ is $\epsilon$-good on $[a, \beta] = [a, \alpha - \delta_{0}]$.

You also need to understand why $f$ is $\epsilon$-good in $[\alpha - \delta_{0}, \alpha + \delta_{0}]$. This is simply because of the choice of $\delta_{0}$ such that $|f(x) - f(\alpha)| < \epsilon / 2$ for $|x - \alpha| \leq \delta_{0}$. Such a choice of $\delta_{0}$ is possible because $f$ is continuous at $\alpha$.

While studying proofs in analysis it is very important that one understands the definitions completely without any ambiguities and it is also important that one tries to formulate definitions with minimal amount of mathematical symbols.

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There exists some $x$ such that $\alpha - \delta_0 < x \leq \alpha$ and $x \in A$. Otherwise, $\alpha$ would not be the least upper bound of $A$.

Because $f$ is $\epsilon$-good on $[a,x]$, and $[a,\alpha - \delta_0] \subset [a,x]$, it follows that $f$ is $\epsilon$-good on $[a,\alpha - \delta_0]$.

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First of all, note that if $x \in A$ and $a \leq y \leq x$, then $y \in A$.

Now if $\alpha$ is the least upper bound of $A$ and $a \leq y < \alpha$, there exists some element $x \in A$ between $y$ and $\alpha$ (otherwise $y$ would be an upper bound for $A$), which shows that $y \in A$ by the above remark. Applying this with $y = \alpha - \delta_0$ shows that $f$ is $\epsilon$-good on $[a, \alpha - \delta_0]$.

For the second part, the same argument works, because at that point of the proof we have already shown that $b = \alpha$.

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  • $\begingroup$ @ Mikko your answer only considers the size of $y$ relative to $\alpha$ to determine whether it belongs to $A$ but in order to be in $A$, $f$ needs to be $\epsilon$-good on $[a,y]$ too. Therefore, does this imply that if we know that $f$ is $\epsilon$-good on $[\alpha-\delta_0,\alpha+delta_0]$, we know that $f$ will be $\epsilon$-good from any number less than $\alpha-\delta_0$ to $\alpha-\delta_0$? $\endgroup$ – mauna Aug 20 '13 at 17:46
  • $\begingroup$ Let me phrase this differently. If $a \leq y < \alpha$, then $y < x \leq \alpha$ for some $x \in A$ since $\alpha$ is the least upper bound. Thus $f$ is $\epsilon$-good on $[a,x]$, which implies that $f$ is $\epsilon$-good on $[a,y]$ since $y \leq x$. Hence $y \in A$. I'm not sure what you mean with that last part. $\endgroup$ – Mikko Korhonen Aug 20 '13 at 18:00
  • $\begingroup$ I've edited my question to elaborate more on the specific area that confuses me. $\endgroup$ – mauna Aug 21 '13 at 14:30
  • $\begingroup$ Maybe what you're missing is the fact that $\alpha$ is the least upperbound. This means that as close as you want to, but below, $\alpha$, there must be an element of $A$. In particular, between $\alpha-\delta_0$ and $\alpha$ there must be an element of $A$, called $x$ by Mikko Korhonen in the comment above. $\endgroup$ – Magdiragdag Aug 21 '13 at 15:01

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