« first day (847 days earlier)      last day (4189 days later) » 

10:00 PM
@N3buchadnezzar I'm trying now ...
@N3buchadnezzar but the name is not actually important
 
No, but I want some more information =)
 
@N3buchadnezzar like what?
 
Who discovered it, similar functions etc.
 
maybe do a serch for non-riemann integrable?
 
I guess it is a proof that a limit can still exist, even if the function is not continuous anywhere! I find that really neat, quite a cool function.
2
=)
 
10:04 PM
no
 
@N3buchadnezzar yep
 
what limit?
 
@N3buchadnezzar I once read a book by Riesz and Nagy which had some excellent discussions of discontinuous/continous functions along those lines
 
@cassandra0 $$ \lim_{x \to 0} f(x) = 0 $$
@OldJohn Thomae's function?
 
it's continuous at 0
 
10:07 PM
@N3buchadnezzar not sure
 
How can a function be continous at a single point?
 
what is the definition of continuity?
 
@N3buchadnezzar easy - $\lim f(x) = a$ at only one point
 
Yeah, it just seems counter-intuitive.
 
@cassandra0 depends - probably sequential continuity in this context
 
10:09 PM
f is cont at a if f(a) = lim {x->a} f(x)
 
@N3buchadnezzar I find continuous non-differentiable functions even more counter-intuitive :)
 
^^ Indeed
 
a limit can not exist if a function is not continuous there
 
@cassandra0 what??
 
it's a tautology from the definition
> The function f is continuous at some point c of its domain if the limit of f(x) as x approaches c through the domain of f exists and is equal to f(c).
 
10:18 PM
@cassandra0 the function which is zero everywhere except for taking the value 1 when $x$ is zero is discontinuous and has a limit at 0
 
you are now talking about a different function
 
Functions, functions everywhere
 
it doesn't have a limit at 0
 
@cassandra0 Oh?
 
$$g(x) = \left\{ \begin{array}{cccl} 1 & \text{if} & x & \text{is irrational} \\ 0 & \text{if} & x & \text{is rational} \end{array} \right.$$
$$f(x) = \left\{ \begin{array}{cccl} x & \text{if} & x & \text{is irrational} \\ 0 & \text{if} & x & \text{is rational} \end{array} \right.$$
 
10:20 PM
@cassandra0 my function has a perfectly good limit at zero - it is just not equal to the value of the function there
 
f is cont. at 0. g is discont. everywhere.
 
And my function is $f$ =)
 
@OldJohn, lim_{x -> 0} g(x) is not defined since there exist two different sequences (one rationals, one irrationals) that converge to different values
 
user19161
I finally got a star after many days with no stars.
 
@cassandra0 yes - I am talking about the function I gave earlier - as a counterexample to your statement "a limit can not exist if a function is not continuous there"
 
10:22 PM
which one?
 
@WillHunting where?
 
user19161
@N3buchadnezzar See the wall dude.
 
"the function which is zero everywhere except for taking the value 1 when x is zero is discontinuous and has a limit at 0" <- this is g no?
oh I see
 
@WillHunting Ah! But I have two
 
I understand, thanks
 
10:42 PM
darn - gone quiet here - was it something I said?
g'night all
 
@OldJohn Nighty mate
That almost sounded like your age too.
2
 
11:13 PM
Good night guys!
 
11:54 PM
I spend too much time answering questions that no one reads.
 

« first day (847 days earlier)      last day (4189 days later) »