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

So I want to prove that continuity on $[a,b]$ implies uniform continuity with only using the least upper bound property of the reals. I know the basic idea of this, but am getting confused with choosing the right $\delta$. Here's where I am so far:

Proof. Let $\epsilon >0$ and define $$A(\delta) = \{u \in [a,b] ~| \text{ if } x,t \in [a,u] \text{ and } |x-t| < \delta, \text{ then } |f(x)-f(t)| < \epsilon \},$$ and $$A = \bigcup_{\delta >0} A(\delta).$$ Since $a \in A$ and $b$ is an upper bound for $A$, $\alpha = \sup(A)$ exists. Now I need to show two things: first that $\alpha = b$, and then that $\alpha \in A$. To show $\alpha = b$ assume that $\alpha <b$. Then by continuity there exists some $\delta(\alpha) > 0$ such that if $|x-\alpha|<\delta(\alpha)$, then $|f(x)-f(\alpha)|< \epsilon$. Now since $\alpha = \sup(A)$, there exists some $x_0 \in A$ such that $\alpha - \delta(\alpha) < x_0 \le \alpha$. Then there exists some $\delta(x_0)>0$ such that $x_0 \in A(\delta(x_0))$. Now let $\delta_{\text{min}} = \min\{\delta(x_0), \delta(\alpha)\}$...

So here is where I am stuck. For starters I'm not sure if this $\delta$ will work. Also, I am imagining that I will need to use the triangle inequality to show that $\alpha \in A(\delta^*)$ where $\delta^*$ is whichever $\delta$ that will do the trick, but I'm not sure what to use the triangle inequality on. Basically I've confused myself. Help?

• math.stackexchange.com/questions/110573/… Sep 20 '15 at 23:46
• @David Yeah I know other ways of proving this and have proved it other ways before, but the point here is to only use the least upper bound property as stated in the first sentence. The choice of delta in that link comes from the compactness of the domain which is not too helpful here. There is a proof that only relies on the LUB property and I am trying to figure that out. Thanks though. Sep 21 '15 at 0:47

I came across this problem when I reading this undergrad analysis book, and I think this problem is a little bit challenging unless I was overdoing it. Here is my proof:

First we have $A=\bigcup_{\delta>0} A(\delta)$ is bounded above by $b$ by the construction of $A(\delta)$, and it's nonempty because it contains $a$. So there exists $c=l.u.b.(A)$, of course $c\leq b$ since $b$ is an upper bound.

Some fact: (1) if $u \in A(\delta)\subset A$, then for any $v\in[a,u)$, $v$ is also in the same $A(\delta)$, therefore in $A$. This is just saying that if there is any $u\in A$, then $[a,u]\subset A$. Proof: play with the definition.(omitted)

(2)For fixed $\epsilon$, if $s\in A$ where A is induced by $\epsilon$, then $s\in A$ where A is induced by $\frac{\epsilon}{2}$. Proof: rescale $\epsilon$.

Since function $f$ is continuous at $c$, for any $\epsilon >0$,there exists $\delta_0>0$ such that for any $y$ that satisfies $|y-c|\leq\delta_0$ we have $|f(c)-f(y)|\leq \epsilon$. Now we pick this very $\delta_0$, since $c$ is the l.u.b.(A) so there exists $s\in A$ such that $c-\delta_0\leq s\leq c$, since $s\in A=\bigcup_{\delta>0}A(\delta)$, so there exists $\delta_1$ such that $s\in A(\delta_1)$. Now we want to show that $c$ is also in $A$:

Fact (1) tells us $[a,s]\subset A(\delta_1)$, let $\delta=min\{\delta_0, \delta_1\}$. Then for any $x,t \in [a, c]$ with $|x-t|<\delta$,

case1: If $x,t$ are both in $[a,s]$, then $|f(x)-f(t)|<\epsilon$ because $f$ is uniformly continuous on $[a,s]$.

case2: If $x,t$ are both in $[s, c]$, then $|f(x)-f(t)|\leq |f(x)-f(c)|+|f(c)-f(t)|<\epsilon + \epsilon=2\epsilon$.

case3: If one is in $[a,s]$ and the other is in $[s,c]$. WLOG, let $x\in [a,s]$ and $t\in [s,c]$, then $|f(x)-f(t)|\leq |f(x)-f(s)|+|f(s)-f(c)|+|f(c)-f(t)|\leq \epsilon + \epsilon +\epsilon =3\epsilon$.

Anyway, we can use fact (2) to rescale and get a uniform $\epsilon$. So we've shown $c$ is also in $A$. Now if we assume $c<b$, then we can repeat the above argument to show $c+\delta_0$ is also is also in $A$ (use $c$ as the intermediate point), contradicting the fact the $c$ is the l.u.b.(A). So the only possible way is that $c=b$ and $c+\delta_0$ is out of the bound of domain $[a,b]$. Therefore $c=b\in A$, done.

P.S.: I've carefully checked my reasoning and it should be right up to some detail work, e.g. rescaling issue. And I'm pretty sure there is a better way to put all these together nicely, but I'm out of time so...