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Is the highlighted part wrong?

were trying to show that the function $f(x) = \tan x$ on $[0,\frac {\pi} 2)$ is not uniformly continuous by using the fact that "if $f$ is uniformly continuous on a bounded set $S$, then $f$ is a bounded function on $S$."

What I don't get is both quizlet and chegg are claiming that $[0,\frac {\pi} 2)$ is a bounded set, when it isn't.

I showed that $f(x) = \tan x$ is an unbounded function on $[0,\frac {\pi} 2)$ hence $f(x)$ cannot be a uniformly continuous on a bounded set $[0,\frac {\pi} 2)$. Yet $[0,\frac {\pi} 2)$ isn't bounded? I don't get this.

Seeker
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kenko
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  • They’re hilighting $[0,\pi/2)$, no mention of infinity? – FShrike Sep 05 '22 at 06:35
  • yeah just fixed that mb – kenko Sep 05 '22 at 06:35
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    You might be confusing boundedness with closedness (or openness). Boundedness is just about whether the set goes off to infinity or not. – Ravi Fernando Sep 05 '22 at 06:45
  • oh ok i get the definition now. I just rechecked the completeness axiom 4.2 def from textbook and that checks out. on wikipedia it says "bounded intervals are bounded sets" but a bounded interval must be upper and lower bounded. so its a unbounded interval hence unbounded set. So a unbounded interval != unbounded set right? – kenko Sep 05 '22 at 06:48
  • yeah i was getting confused with the closedness or openness. thx – kenko Sep 05 '22 at 06:50

2 Answers2

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A set is bounded if it has both upper and lower bounds, i.e. if there are values $l$ and $u$ such that every element $x$ from the set satisfies $l \leq x \leq u$. The interval $[0, \frac{\pi}{2})$ is bounded - it has a lower bound of $0$ and an upper bound of $\frac{\pi}{2}$. (It has many bounds, in fact, but these are the tightest ones.)

It's true that the interval does not have a maximum element, because its supremum (aka its least upper bound) lies outside the interval itself, but that's a different property.

ConMan
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You have to show that there exists an $\epsilon>0$ such that for each $\delta>0$ there are $x,y$ in $\left[0,\dfrac{\pi}{2}\right)$ such that $|\tan x-\tan y|\geq\,\epsilon$. This is equivalent to finding two sequences $x_{n},\,y_{n}$ such that $|x_{n}-y_{n}|<\dfrac{1}{n}$ and for large $n$ to have $|\tan(x_{n})-\tan(y_{n})|\geq\,\epsilon$.

Take $\epsilon=1$ and $x_{n}=\dfrac{\pi}{2}-\dfrac{1}{n^{2}}$ and $y_{n}=\dfrac{\pi}{2}-\dfrac{2}{n^{2}}$.

It is clear that $|x_{n}-y_{n}|=\dfrac{1}{n^{2}}<\dfrac{1}{n}$.

Now we use a formula of trigonometry: $\tan A-\tan B=\dfrac{\sin(A-B)}{\cos A\cos B}$ and we get :

$|\tan x_{n}-\tan y_{n}|=\dfrac{\sin(1/n^{2})}{\sin(1/n^{2})\sin(2/n^{2})}$

and for $n$ sufficiently large we obtain

$|\tan x_{n}-\tan y_{n}|\geq\,1=\epsilon.$ The fact that $\tan x$ is

unbounded on $\left[0,\dfrac{\pi}{2}\right)$ is obvious!

You don't have to invoke any theorem. This proof essentially includes the theorem!

But if you want a proof of your theorem READ this!

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A uniformly continuous function maps bounded set to bounded sets

Cheese Cake
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