Question

I need help with open sets, how can I demonstrate that { (x,y) | |x|<1 , |y|<1 } is an open set?

Answer 1

You know the definition of open set?

Answer 2

yes...kind of...but that is my problem I know the definition but I really don't get how to make de demonstration :S

Answer 3

Show it here

Answer 4

if you write\[|x|<1\]that is the same as saying that \[x\text{ is in the interval} ~~(-1,1)\]

Answer 5

\[\left\{ X \in R^n | \left| X-Xo \right|<\delta\right\}\]

Answer 6

ah you want something rigorous
my mistake :P

Answer 7

I think the following assertion should be sufficient:
For any (x,y) in the set, there exists a circle with radius min{(1-x)/2,(1-y)/2}

Answer 8

yup, but what is \[X_0\] here?

Answer 9

is an elemnt of the set, right?

Answer 10

yeah, IsaacDian right. Using the defition, \[\forall x \in A\], there exists r>0 such that \[B(x,r)\subset A\]

Answer 11

but now that I have that how is the demonstration??? Is just I get why is an open set but how do I demonstrate that is in fact an open set :S

Answer 12

I think you can see that (-1,1) is an open set then (-1,1)x(-1,1) is also an open set. You can display it on xOy

Answer 13

All of these are demonstrations.

Answer 14

oh! I think now I get it!!!

Answer 15

Then consider the set defined as S:={ (x,y) | |x|<1 , |y|<1 }U{(1,1)}.

Answer 16

Yes IsaacDian thnx I think now what have to write is that because {(1,1)} is an open set then {(x,y) | |x|<1, |y|<1} is also an open set, am I right?

Answer 17

If you want to prove rigorous, you should use IsaacDian's proof.

Answer 18

No, that was a different question...
Notice that {(0,0)} is a singleton, with only one element, namely (0,0), and by U, I meant the union of two sets.

Answer 19

U_U sorry I'm lost with that last reply...

Answer 20

Now I think I'm confused O_O because the circle, I think, should be define like B((0,0),1)