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thoughts?
if you see a strict inequality, that indicates the set is not closed
but double check to be sure
it's certainly not compact, because it is unbounded. E.g we can take y = n and x=1
$$ \left\lvert\frac{n}{\sin (n^3)}\right\rvert \geqslant n $$
@craggy terrace
aL
heuristically i'd eliminate a,b and c
so that pretty much leaves d
verify it's open and you're done
easiest way I can think of is by showing the complement is closed
definitely not
sin is bounded, n^3 is unbounded
you're thinking of sin x ~ x as x->0, not relevant for my argument
I see
do you understand the point of this?
Actually no
But i understands that if it is unbounded then it is open
Not closed
is the pair (1,10) in the set, for example?
this is the condition
actually I mis spoke
,w sin(1000)
the pair (1,n) is in the set for infinitely many n
but it still means the set is unbounded
Yes it is inside
,w sin (8000)
10 something
It still
so take (1,20) then
so the result is positive not negative
i dont remember
sin n^3 is positive for infinitely many n
you should focus your effort in proving the set is open
rather than all this secondary stuff
But it is negative too?
,w sin(10)
And it is negative also for many n no?
it is
sine is positive whenever it is between (0, pi) or (2pi, 3pi) or (4pi,5pi) etc etc
Right
So?
@desert cosmos
+close