Using an appropriate definition of continuity, show that the continuity of f (.) at x is equivalent to
lim h->0 sup|t|<h |f (x + t) ─ f (x)| = 0
Recall that the limit of lim x->a f(x) = f(a) so what you are doing is instead of considering the epsilon delta definition of |f(x) - l| < e where the limit is l, then this limit l is exactly f(a) = l for continuity. If f(x+t) -> f(x) then that expression should be zero since the limit approaches l = f(x) [Recall: the definition of continuity is that the limit is equal to the value of the function).
Is this the statement?
If so, then proof is an proof. To get you started:
Each of those requires an argument, but is very straightforward. So you can see that the final form there is almost identical to what you're trying to prove, except for that sup (supremum = least upper bound) part.
To prove the proposition, you'll need to prove two things, the direction, and the direction.
When you're into the guts of the proof, you'll want to choose (in one direction, though this works for both) the so that
it makes the limit you know
This won't make any sense until you try to prove it yourself (but perhaps then it will be helpful):
If everything in a set is less than , then the supremum of that set will be less than or equal to , which is less than
That will be your decisive observation when proving it in one direction.
In the other direction, you'll observe that if the supremeum of a set is less than (or you could use there too),
then any individual one of those things in that set is less than
hi thanks for your replay, but i'm not getting it, i tried solving it using epsilon-delta method but couldn't complete it another unclear thing is "sup". how to deal with it.
as i'm new to this stuff so i need little more help in solving it. I'll appreciate if you help me to understand it.