Given the properties (1) so that and (2) f is continuous at a if, for every sequence such that , the sequence converges to .
I need to prove . We did the first part in class:
I employ the contrapositive approach. Suppose f does not have property (1) and suppose that . Now if (1) does not hold, then there is so that, , . When . So in there is some so that
Now I am to my problems: 1)I need to prove that, in fact, . Let . I need to find so that, for .
I am inclined to let , but then might not be an integer. So I am not sure what to do about that. Once I find my , I simply need to finish the epsilon proof of this part.
2) I need to prove that, in fact, . This means that it is not true that so that, for , then . Since when , it seems like I am done, but I am not sure how to make the conclusion.
Sorry this post is so long, but this is a long problem. Thank you for your help!
If you want to prove 2 1 , then consider this
is an open set in X for every open set in Y because f is cont
contains a and is an open set of X
open neighbourhoods of a, i.e r >0, , where S is the set containing ; actually is eventually in
so is eventually in