Sorry for the delay, @Hippa. The key is knowing the topology of $[0,1]$; you can't really do this otherwise. His proof and Daniel's are basically the same thing. Pick a path between any two points $x \in U$, $x \in U^c$. Consider the set of points $f^{-1}(U)$; convince yourself it's open. Then its complement is closed. In $[0,1]$, this means it has a least element: a smallest $c$ such that $f(c)\in U^c$. This is your limit point.
Daniel's proof would go as follows: pick a path $f$ connecting $x$ and $y$. Then $f^{-1}(U)$ and $f^{-1}(U^c)$ are both open. The unit interval is connected, so one of them must be empty; contradiction, there was either no $x \in U$ or no $y \in U^c$.
Sorry. It's our path $[0,1] \to V$.
So understanding this situation gets pushed back to understanding the topology of the reals.
With the definition... let $f$ be continuous. Pick a point $x \in f^{-1}(U)$. By definition $f(x)\in U$. Pick $\varepsilon$ small enough that $B(x,\varepsilon)\subset U$. Then by definition of continuity...
@r9m How about this one? $\displaystyle\frac{\displaystyle\left(1+2\sqrt{4!+5!}\right)\left[\prod_{n=1}^3\Gamma\left(\frac{n}3\right)\right]^2} {\displaystyle{e}^{2\operatorname{Li}_1(\frac12)}\operatorname{Li}_2\left(\frac{\Gamma'(1)}{\Psi(1)}\right)}=50$
@teadawg1337 Thanks. I had been working with lots of product identities at the time and so the proof just seemed obvious. I've found that what you are working on at the time greatly affects the proofs you come up with
@teadawg1337 I don't know any closed form for $\Gamma\left(\frac13\right)$ or $\Gamma\left(\frac23\right)$, but I know their product :)
@teadawg1337 For $n=3$ we can use the reflection formula, too
i cut a mobieus strip that had one 180 degree turn in it, down the middle, and I got a longer one, then two turns and i got two rings, three and i got a knot, and four i got two knots, will i just keep getting knots if i keep going? i ran out of paper
@r9m but for $n$ turns, $n$ even i have two surfaces, and $n$ odd I have one, but $n=1$ i have a middle cut giving me a longer mobieus strip, and $n=2$ giving me two interlinked mobeius strips, but $n=3$ gives me a knot, and $n=4$ seems to give me a knot too!
I'm not sure if this is the right place to do this, but could a moderator (or anyone) explain why this comment is acceptable on this site
" @Behaviour Im sorry but I think you are being not nice. Maybe she has more time " from now on " or something like that. Surely she is an intresting user of MSE. Or maybe Im just defending her because she is pretty :) But still I think she makes a good candidate." math.stackexchange.com/election/5#post-1059198
I don't want to call anyone out publicly, but I flagged that comment and the flag was rejected.
@copper.hat I was also expecting that they would have a conversation with that person. Looking at his profile, he sais that he's ~15 years old, so that probably is a factor. I would hope that they would explain to him the right way to participate in an online community.
@copper.hat You said "that would cover a lot of comments", and I was throwing in the other adjectives, too. I'm not suggesting it should be deleted because it's not particularly constructive on the nomination, I'm suggesting it should be deleted because, as you said it's trite, borderline misogynistic, say more about him than anybody else, and as I said, serves no other purpose.
I don't see why it should stick around.
Anyway, this doesn't concern me, so I'll stay out of it.
@MikeMiller: thanks! i suspect censoring the comment would generate distracting flak. anyway, i should stay out too, i was just curious. thanks for helping earlier, much appreciated (turns out that i misunderstood the system response, my paragraph was too long, i just needed to split the paragraph, not get someone else to interject.)
as an aside, i think mods should have some minimum rep as a proxy for maturity of some sort. also lets one evaluate by looking at their responses, etc. (that said, the acrimony that has appeared in the past between high-rep mods suggests my though process is flawed.)
it maps to \mathbb{R}^{n-k} and i was expecting to see the range to be R^n
hi ted!
excuse my pedestrian typing
@MikeMiller: i don't need to know about the whole post, just if there is some 'dual' way of expressing a patch/coordinate thing in manifolds that i haven't got to yet?
usually it is something like $\phi:U \to \mathbb{R}^k$ or the like, here they have $n-k$ instead of $k$.
:-). i used to have a collection of friends/colleagues with whom i could ask trivial questions, but that was many years ago and they have very dispersed time zones...
and i trip on very basic stuff, not so much the technical content as the underlying idea.
@TedShifrin i was visiting my parents in their rental cottage. They come down for a couple months every year to see me, and because it's Florida and they're retired.
Anyway, @Christofian, you don't seem to be here, but mick was not being blatently rude, and although the part of the comment you highlighted is irrelevant, the comment on the whole is not. One could read into his intentions in many different ways, but when it comes down to it, none are offensive enough to warrant moderator intervention.
whereas a lot of the ideas of point-set topology are about local phenomena, coarse geometry is tailor-made to only notice the difference at global levels
two things are topologically the same if you can squish one into another; two things are coarsely the same if you can change the local data from one to the other; $\Bbb Z$ is coarsely equivalent to $\Bbb R$
Also, Remmert talks about valuations on fields of meromorphic functions when he means discrete valuations, which makes me wonder if there are non discrete valuations on such fields. He shows the only discrete valuations on those fields are the order valuations times integers.
@MikeMiller Why does the guy speak of "...**the** fundamental group."?
If I'm given the graph of f'''(x), how can I determine the points of inflection of f(x). Would they be located at x values where f'''(x) attains a max or min? Or what?
(I'm not 100% sure because I haven't done integrals yet, I just know what they do.) It's easier to visualize with f'''(x) being velocity as a function of time: v(t) and f''(x) being distance (displacement?) as a function of time: s(t)
The area under v(t) is distance, so when that area is 0, s(t) = 0, so t is a root.
Suppose $X$ is a connected space with the property that every nondegenerate closed connected subset is disconnected. Then is every proper subcontinuum of $\beta X$ nowhere dense?
@TomCruise Consider a solid torus. Call it $T_0$. Take a torus wrapped up twice inside it. Call it $T_1$. Take another torus wrapped twice inside $T_1$. Call it $T_2$. etc...
Now consider the space $\bigcap T_i$
That is called the solenoid (the geometric topologists interpretation).
My intuition comes from a different viewpoint though. The solenoid can be thought of as a nontrivial fiber bundle of S^1 and the cantor set, and the cantor set has the property you want.
Discussion between copper.hat and use…
Discussion between copper.hat and use…
