Recent content by marlen

S = {(x,y): y = 1/x, 0 < x \leq 1} \cup {(x,0): -1 \leq x \leq 0} I meant would this work?

TOTALLY DISREGARD THIS COMMENT

S = f(x; y) : y = 1/x; 0 < x  1g [ f(x; 0) : 1  x  0g would this work

Huh? I'm confused. Would 1/x work? What would be a solid example, one that I could understand?

Connected/Disconnected...all the same to me... My question for all of you ladies and gentlemen is what would be considered as an example of a connected set in R squared that becomes disconnected when we remove one point. My answer would be sin(x/2), but is there a simpler example.

Hello, Can someone please help me prove that the lim as n goes to infinity of (the sequence an + the sequence bn) = (the lim of an) + (the lim of bn). What I think is that if one adds the two limits an + bn, she will come up with a new sequence cn and take its limit, which equals L...

Homework Statement Let S be the nonempty set of real numbers bounded above. Prove that S^3 = {x^3 : x \in S} is bounded above and sup S^3 = (sup S)^3 Homework Equations given S^3 = { y \in R : \exists x, x \in S and y = x^3} and for all \epsilon > 0, there is y \in S^3 such that...

Thank you arildno and everyone else who helped us. It makes sense now! ;)

Yea, but you can't just say that. I know it's true but how do i show that.

but we can't just assume that there lies a rational or irrational number in the the interval of (x-r, x+r) can we?

I'm not really sure how to answer this question.

i suppose but does that matter

no. ... so what you are saying is that if we have an irrational number x, and for some small r>0 we get the interval (x-r, x+r) which contains either a rational or irrational number, meaning we cannot draw an open-ball around our x, meaning that the set of irrationals cannot be open.

The complement of irrational numbers is the set of rational numbers. and this set can't be closed or open, i think...because the real numbers is an open set and i know that the union of two open sets is open. but irrational numbers are not open, so I'm stuck

No, the interval will contain either another irrational number or a rational number.