1. ## boundedness

may i know why "if the function f on a closed bounded interval [a,b] is unbounded then there exist a seq x_n in [a,b] st l f(x_n)l > n for all n"

i was trying to list a few examples to show why this is true but i do not get it.

it seems to be that the function f on a closed bounded interval [a,b] can be bounded and yet there exist a seq x_n in [a,b] st l f(x_n)l > n for all n.
this was my working:

assuming that the function is from [0,10] where the range is [ 20,30] which means that f(x)= 20+x

then taking the even seq, f(x_1)= 22, f(x_2)=24... where 22>1, 24> 2.. hence
f(x_n) > n for all n, where in this case, the function is bounded.

may i know what mistake i have made?

thank you!

2. In your proof what do you pick for x_40?

3. Originally Posted by alexandrabel90
may i know why "if the function f on a closed bounded interval [a,b] is unbounded then there exist a seq x_n in [a,b] st l f(x_n)l > n for all n"

i was trying to list a few examples to show why this is true but i do not get it.

it seems to be that the function f on a closed bounded interval [a,b] can be bounded and yet there exist a seq x_n in [a,b] st l f(x_n)l > n for all n.
this was my working:

assuming that the function is from [0,10] where the range is [ 20,30] which means that f(x)= 20+x

then taking the even seq, f(x_1)= 22, f(x_2)=24... where 22>1, 24> 2.. hence
f(x_n) > n for all n, where in this case, the function is bounded.

may i know what mistake i have made?

thank you!
Think of why counterexamples won't work. But think about it this way. Suppose that the process of recursive definition that's going on here. Clearly there exists some $\displaystyle x_1$ such that $\displaystyle |f(x_1)|>1$ otherwise $\displaystyle |f(x)|<1$ for all $\displaystyle x\in[a,b]$ which contradicts boundedness. Suppose then that you keep doing this and you come to some $\displaystyle n\in\mathbb{N}$ such that there does not exist $\displaystyle x\in[a,b]-\{x_1,\cdots,x_{n-1}\}$ with $\displaystyle |f(x_n)|>n$ then for all $\displaystyle x\in[a,b]-\{x_1,\cdots,x_{n-1}\}$ one would have that $\displaystyle |f(x)|\leqslant n$ and thus surely $\displaystyle |f(x)|\leqslant \max\{f(x_1),\cdots,f(x_{n-1}),n\}$ for all $\displaystyle x\in[a,b]$ contradicting unboudness too!