Will every subset of $R$ that is not bounded above contains a sequence that diverges?

Question

Question:

(a) Prove that every subset of $R$ that is not bounded above contains a sequence that diverges to infinity.

(b) Prove that every unbounded subset of $R^d$ contains a sequence ($x_n$) with the property that $\lim_(n_\to ∞_)$ ||($x_n$)|| = $∞$

Idea:

Consider a non-empty subset $A$ and a sequence ($a_n$). For part a), since the subset is not bounded above, then

there exist a number t such that $a ≥ t$ for all a $\in$ $A$.

the supremum of $A$ does not exist.

it diverges to ∞ and the $lim sup$ ($A$) does not exist as well.

These are the things that I can conclude from the given, but I am not 100% sure that all of them are correct. I am trying to find some connection between the subset and its sequence, so that I can prove since the subset diverges to infinity, then the sequence also diverges to infinity. However, I am not sure if I am in the right track. Should I prove this by contradiction?

For part b), I started by listing the definition as well, but I don't know how to continue after I list the definitions.

Any help will be super appreciated, Thanks

Let me know if the context is not readable.

Answer 1

The trick is to use the integers, which we know diverge to infinity. If $A$ is not bounded above, we construct a sequence $(a_n)$ with $a_n\in A$ for all $A$ such that $\lim_{n\to\infty}{a_n}=\infty$. We ensure this by constructing the sequence such that $a_i>i$ for all $i$. This can be done independently for each $i$: given that $i$ is not an upper bound of $A$, there exists an $x\in A$ such that $x>i$; set $a_i=x$. (This is really invoking the axiom of choice since there are infinitely many arbitrary choices).

Answer 2

For (b):Let $f(1)=1$. Let $f(n+1)>f(n) $ such that $|a_{f(n+1)}|>1+|a_{f(n)}|.$ By induction on $n$ we have $|a_{f(n)}|\geq n+|a_1|.$ For (a):Let $f(1)=1$ and let $f(n+1)>f(n)$ such that $a_{f(n+1)}>a_{f(n)}.$ By induction on $n$ we have $a_{f(n)}\geq n+a_1.$