Conversation started Oct 5, 2016 at 4:04.
Oct 5, 2016 04:04
Let X be a metric space, let A be a non-countable subset of X s.t there exists E > 0 for all $a,b \in A$ one has p(a,b) >= E. Then X is non-seperable.
cool @0celo7
@Adeek p?
metric
what's wrong with d :o
I dunno prof likes his p's
haha
you sure it's a p not a rho
aha
I have found the uniform continuity proof
I don't think I need Lebesgue for the homework, but I should point out he told us a wrong proof
Oct 5, 2016 04:30
oh
yeah
I think you should prof like it as well when you do that
Oct 5, 2016 05:27
Is (a,b) open in all (R,d)?
arbitrary d such that (R,d) is met space
R being the reals
use a bijection on R that sends (a,b) to a set that is not open wrt the Euclidean metric. transport said metric using the bijection.
(if f:R->R is such a bijection, define d as d(x,y)=|f(x)-f(y)|)
roger that
hey @arctictern would you like to discuss the problem above with me ?
nah. not in an analysis mood, youtubing off to sleep.
oke
Oct 5, 2016 05:43
@Adeek You can find disjoint balls with centers in the points of $A$. You have uncountably many of them. If $D$ is dense in $X$, then each of this balls contains at least one point from $D$. So $D$ is uncountable.
I guess this is quite a common question so that are probably several posts about his on the main site.
good thing that your here @MartinSleziak I am struggling.
I simply put uncountable separable metric "d(x,y)" site:math.stackexchange.com into Google. Perhaps if you try searching for some different - but related - phrases, you will find other questions about the same problem.
oh I see
assuming that result you posted above. if I consider for example $l_p(\gamma)$ how can I find f,g that satisfy this condition ? I mean the domain that we are talking about is arbitrarily uncountable set ?
BTW there is also general topology room. Your question would fit nicely in the topic of the room. However, for quite a long time there is almost no activity in that room, so it might go unnoticed if you post there. But let's hope that the general topology room will become active later again.
oh awesome I will go there
This is my question in case you didn't see it.
Oct 5, 2016 05:51
So we are talking about norm $\|x\|=\left(\sum_{i\in I} x^p\right)^{1/p}$ where $I$ is a set of uncountable cardinality.
yes
Ok, in your notation you have $\Gamma$ instead of $I$.
yeah
yes
@Adeek Wouldn't such functions which have 1 exactly at one position and zero at others work?
I.e., I define $f_\gamma$ by saying $$f_\gamma(x)=
\begin{cases}
1 & x=\gamma, \\
0 & x\ne\gamma.
\end{cases}$$
Sorry, Adeek, that's basically all what I can do now. I will have to leave soon - time to go to work. :-(
hm I will think about it thank you though
I dunno why people
didn't like my question in main
Oct 5, 2016 05:54
BTW I consider the space $\ell_p(\Gamma)$ a nice thing.
It is interesting to know that we are able to work quite nicely with uncountable sums.
yeah
I like that as well I generally like infinite uncountable spaces.
it is very nice thing to stretch imagination.
yes
@MartinSleziak defining it this way we get actually what we want
because if we lets say have f as you defined it
And also I like the result that each Hilbert space is isomorphic to $\ell_2(\Gamma)$ for some $\Gamma$. (Although I do not know about some reasonable application of this results. But still, knowing that they are completely characterized and that that there is one simple cardinal invariant which determines whether two such spaces are isomorphic is kind of nice.)
and we define another g in same way where $\gamma$ is chosen in a different than the other one that you did, then we get $d(x,y) = \alpha$
or I guess $d(f,y) = \alpha$ under that norm.
Here $\alpha=2^{1/p}$, right?
in particular $\alpha = 2^{1/p}$.
yeah
yes so that works.
also for $l_{\infty}(\Gamma)$ we can do same argument actually
@MartinSleziak I really like functional analysis
it is nice
@MartinSleziak before you go do you have an idea about which functions to use in $c_0(\Gamma)$?
Oct 5, 2016 06:05
@Adeek My guess would be that the same functions should work.
ok, I'll have to go, see you later
alright cya l8er
thanks a lot for your help @MartinSleziak
yes your righttttt
same function work
thanks a lot @MartinSleziak
 
Conversation ended Oct 5, 2016 at 6:09.