the set {1}R++ is open close or both
or neither

Question

the set {1}R++ is open close or both
or neither

anonymous · Answer

what is R++

anonymous · Answer

set of Real nos in 2D

anonymous · Answer

{1} will be closed

anonymous · Answer

any set with finite number of elements is closed in R^2

anonymous · Answer

when I say it is closed I asume you using euclidian metric, not descreet.

anonymous · Answer

@myko can plz u elaborate!!!

experimentx · Answer

i don't understand how you are representing {1} as subset of $ \Bbb R^2 $ shouldn't it be tuple of two spaces?

anonymous · Answer

good point, :) @experimentX

anonymous · Answer

experimentX is right . My bad. {1} does not belong to $R^2$. Because this set is obtained as product R x R

anonymous · Answer

so it should be maybe { (0,1)} or {(1,0} and simillar

anonymous · Answer

hav no idea abt euclidian metric..
and @experimentX ur point is rite bt tis wat was given in que

anonymous · Answer

so then you should say maybe that {1} does not belong to R^2

experimentx · Answer

you sure about it? I am not so good with set theory. When you prove that set is closed or open ... you have to go through the standard definition.  A set if closed it it contains it accumulation point. A discreet point (in euclidean space where the distance between to points is defined by the usual Pythagorean distance), the discreet point is a closed set. But I am not so sure ... how a coordinate point is represented by a single element. But i've heard of it.

experimentx · Answer

" when he came across a theorem which stated that points in the plane could be specified with a single coordinate" http://en.wikipedia.org/wiki/Wac%C5%82aw_Sierpi%C5%84ski maybe i misunderstood it.

anonymous · Answer

@experimentX  thanks man!  n m sure abt d que...

anonymous · Answer

If it's true that the R^2 element can be asociated with an element of R, which I strongly doubt, it would be closed.