Prove that the closed interval [0,1] is a closed set and that the open interval (0,1) is an open set..
Could I just say that for [0,1], every open ball B(0,r), r > 0 contains at least one point less than 0 and therefore not an element of [0,1] to prove it is closed
and for (0,1), its complement (-inf, 0] U [1, inf) must be closed by similar reasoning?
is there a way to incorporate the theorem:
a set S is closed iff for all sequences {x_{k}} such that x_{k} is an element of S for each k and x_{k} --> x, it is the case that x is an element of S
I feel like Im basically just stating the definition of closed/open sets up top, and I don't know if that's an acceptable way to do it
Thanks
Sorry I'll try to write it more formally.
Using this definition of an open set: A set S in R^{n }is open if for every element x in S, there is an r > 0 such that B(x,r) is a subset of S
suppose [0,1] is open
[0,1] contains the element 0
B(0,r) contains (0 - r) which is not an element of [0,1] for any r > 0
therefore B(x,r) is not a subset of S for all x in S, which implies [0,1] must be closed by contradiction
No, I thought that's what was being asked here though?Do you know how to prove sets of real numbers are open or closed?
Im interested in the proofs you suggested, could you explain them to someone who hasn't taught the material for 40 yrs? I don't understand how that intersection being empty says anything about [0,1] being closed. I could replace [0,1] with (0,1) and the statement would still be true..