Can someone explain why in game theory it is never correct to choose a strategy that isn't a best response to some belief about what the other player will do?
My son is in 2nd grade. His math teacher gave the class a quiz, and one question was this:
If a triangle has 3 sides, and a rectangle has 4 sides,
how many sides does a circle have?
My first reaction was "0" or "undefined". But my son wrote "$\infty$" which I think is a reasonable answ...
@PM2Ring Topologists answer is definitely bonkers here. There's different kinds of coffee cups, some would have one, some would no holes, some could have two. The philosopher's answers is a close second, but chemist killed me, lol.
I've been comfortable with infinite sets, limits, etc, for decades. But I wouldn't claim that I really understand any of the various infinities that we use in mathematics.
How about we define a vertex as a non-differentiable point of the boundary. Then it'd have no vertices. Then we could define a side as a smooth curve between two vertices.
OTOH, I think a "simple" infinite entity like the set of counting numbers is easier to grasp mentally than some huge finite number. I can write 1000!, but can I honestly say that I really appreciate its magnitude?
@Jakobian Ok, I guess that works. Sort of. ;) But it means that a circle has the same number of sides as the empty universe, which is a bit unsatisfactory.
Hello. I have a dummy question: is the n-dimensional sphere without the north-pole $S^n\setminus\{N\}$ an open set in $\mathbb{R}^{n+1}$?
Let $S$ be the south pole. $S^n\setminus\{N\}\cup S^n\setminus\{S\}=S^n$ why does the union of two open sets give a compact (thus closed) set? What is the problem? Maybe the punctured sphere is open only with respect to the topology induced on the sphere a regarding the sphere as a topological space, it would be an open set as well
I don't see where you answered the question of $S^n\setminus\{N\}$ being open in $\mathbb{R}^{n+1}$. I see that you have postulated that $S^n\setminus\{N\}$ is open in $S^n$ with the subspace topology, which is a correct and good observation, but not an answer to the first question.
Oh, okay. I asked the first question as a preamble for the second one that was my actual question. Regarding the question about $S^n\setminus\{N\}$ being open in $\mathbb{R}^{n+1}$, I would say no. If $S^n\setminus\{N\}$ and $S^n\setminus\{S\}$ were open sets in $\mathbb{R}^{n+1}$ their union being a closed set (i.e. the sphere) would be absurd
What about the open sets $\mathbb{R}^n\times(-\infty,1)$ and $\mathbb{R}^n\times(-1,\infty)$ in $\mathbb{R}^{n+1}$? Is their union not closed in $\mathbb{R}^{n+1}$?
@Thorgott in that case, though, the union is the entire topological space that by definition is closed and open at the same time (together with the empty set)
@Feynman_00 In $\mathbb{R}^n$, the only clopen sets are the empty set and the entire space. Hence the union of open sets will not be closed unless it is the entire space. Other spaces permit more interesting examples of unions of open sets which are closed (or compact, even).
The example of $S^n \setminus \{N\} \cup S^n \setminus \{S\}$ is a good example, in the subspace topology on $S^n$.
Or give the integers the discrete topology. Every set is clopen, and every finite set is compact.
The first point you raise is clear to me. Regarding the second point (the one about other spaces), I don't see how the union of a countable number of open sets can be closed (except they cover the entire space or they all are the empty set)
@Feynman_00 Consider the space $\mathbb{R}\setminus \{0\}$. Cover $(0,\infty)$ with a countable collection of open intervals.
The union of those intervals is closed in $\mathbb{R}\setminus\{0\}$, and is not the whole space.
However, if $A = \bigcup_\alpha U_\alpha$ is closed, where each $U_\alpha$ is open and the $\alpha$ come from some (possibly infinite, possibly uncountably infinite index set), then $A$ is a connected component (or the union of several connected components).
To see this, suppose that $X$ is the ambient space containing $A$. Then $B = X \setminus A$ is open (since $A$ is assumed to be closed). Then $A \cup B = X$, and $A \cap B = \varnothing$. Thus the space is disconnected, and, in particular, no point of $B$ lives in any of the connected components of $A$.
I'm sorry but I'm weak in topology and I need more time to think about this. How does this not contradict the definition of a topology as a collection of open sets such that, among other things, any arbitrary (finite or infinite) union of members of the topology belongs to the topology?
@Feynman_00 This is true for some spaces, e.g. $\mathbb{R}^n$. But there are all kinds of spaces with lots of clopen sets. Every open set in the $p$-adic numbers is also closed. Every subset of a space with the discrete topology is clopen. Every component of a disconnected set is clopen. And so on.
For the record, while Xander is finishing the point about connectedness, I want to point that we can also observe that $S^n\setminus\{N\}$ is not open in $\mathbb{R}^{n+1}$ directly by definition. To be open would require containing a small ball around each of its point, but actually if you take any ball around any point in$S^n\setminus\{N\}$, it will contain points not in $S^n\setminus\{N\}$, e.g. by moving slightly in the direction perpendicular to the sphere.
Answering to Thorgott, now that I think about it, I used the definition to answer my question about the punctured sphere a few days ago, then the problem arised with the second question but now everything is clear. Thanks.
Here is a different proof from the ones in the comments.
Let $O=(0,\dotsc,0)$. If $\mathbb{R^n}$ and $H^n$ were homeomorphic, then let $P$ be the image of $O$ in this homeomorphsim. Then $\mathbb{R^n}\setminus\{P\}$ and $H^n\setminus\{O\}$ would be homeomorphic, too. In particular, they would be...
Are the following definitions of relative countable compactness equivalent?
$\overline{A}$ is countably compact
Every sequence in $A$ has a convergent subnet
Note that for regular spaces, the analoguous definitions are equivalent for relative compactness. Also, the equivalence always holds if...
@Jakobian It was not closed for no reason. It was closed for the reason
> Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
"The reason doesn't make sense to me" is different from "It was closed for no reason."
I will also admit to some confusion about the statement "Every sequence in 𝐴 has a convergent subnet," though it could just be that my topology is rusty: why do you consider a subnet of a sequence? Nets generalize sequences. If you are already working with sequences, why do you ask for the generality of nets?
That being said, I likely wouldn't have voted to close the question, and a polite discussion in CURED might get you enough upvotes to have the question reopened.
(Charging in and saying "My question was closed for no reason!" is not likely to win you support.)
i think a somewhat generous reading of the last two sentences of the post would count as "possible strategies" and/or "your current progress" although this might not be apparent to the average SE user.
they should have warning labels attached to profiles like that. he answered that in 15 minutes. people should have advance warning that they're dealing with someone who's walking around thinking about the deleted tychonoff plank.
ajay: just the usual, working in the day, goofing off at night. it's been hot lately and my daughter makes me take her swimming for at least an hour a day.
ted: i probably had a half-semester window in which i could have written that too. and i've thought about nets. so what's my excuse.
we haven't been to the duck pond in weeks due to the heat. we did see the coyote while checking the mailbox on thursday. and last night i saw two coyotes outside our porch.
All students have to deal with stress to a certain extent. I know my classes stressed out some of my students, but others relished them. I am a worrier by nature, so I do stress, but not usually out of control; there have been a few exceptional times — and I did go talk to a professional for a few weeks during one of those.
Same, in prior months, due to stress from exams and other work, i've become impulsive and am quick to anger. I have been seeing a therapist for a while, and it has been helping.
Well, I guess as we all grow older we find better ways of dealing with stress...
Anyways, enough with the boring talk... How have you been these few months?
I think the best way not to get stressed by exams is to prepare well ahead and not procrastinate. In my 40+ years of teaching experience, I found that the procrastinators put themselves in binds repeatedly.
Hello friends! I've a doubt related to applications of derivative. ___ I was learning about maxima and minima using second order derivative test in which first we differentiate the given function $f(x)$ and then solve for $f'(x) = 0$. Putting the critical points so obtained in $f''(x)$ tells us about maxima and minima... If $f''(x) > 0$ implies x is local minima. While $f''(x) < 0$ implies that x is local maxima. ___ My question is that -- how is second order derivative able to tell whether the point is local maxima or minima. Just like first order derivative can be defined as the rate of …
putting aside where f''(x) = 0 (where the test is inconclusive, but is worth mentioning - this 'test' fails says nothing about even a simple function like f(x) = x^4) one way of looking at it is that the sign of f'' tells you concavity. f'' being positive means the graph is smiling near the point (so if it has an extremum its a minimum). f'' being negative means the graph is frowny so if it has an extremum it will be a max.
you can also frame stuff about f'' simply in terms of the first derivative (of a function that happens to be f'). if f''(c) is positive, this suggests that f' is increasing near c. so if f' is 0 at c, then f' should be negative before c and positive after c. and step back and ask what that tells you about what f is doing. (decreasing and then increasing, i.e. hitting a local min.)
a graph is "smiley" when as you move from left to right the slopes of the tangent lines increase. that means f' should be an increasing function, or as ted would have it (f')' should be positive.
LOL, I was more concerned about my telling you to look at upward- and downward-pointing parabolas. Draw tangent lines and look how their slopes behave as you move left to right.
Most textbooks explain this, and I certainly would hope every calculus teacher explains it carefully.
For several terms, I worked with grad students their first time teaching calculus — discussed how they should present things, worked with them on exams, and watched them teach several times. This was certainly one item we covered.
Ultimately, the department hired someone to do this officially as one of their courses — both for precalculus and for calculus.
It was always amusing dealing with grad students who figured they were going to be the genius that does everything "right" ...
I still shudder when I think about some of the very experienced professors at MIT who lectured multivariable to 300+ students for the first time and just were so bad at it.
@robjohn @leslie @copper @Xander This was a fun comment to write.
No need for that, @HelpMe. But, truly, drawing pictures and seeing things before your very eyes helps more than formulas. Even @leslie will grudgingly agree with me.
Some teachers will agree with me and some will disagree. But when I taught integral calculus and taught different methods of finding volumes, I would not give complete credit unless students drew a sketch and showed me how they thought through the problem. So many students struggle on this because they try to do it by rote memorization and don't understand what's going on.
@Xander Well, having seen some questions from that answerer, I was a bit taken aback by his tone of authority.
@HelpMe Yes, sadly, too many teachers are rigid and don't understand that different students have different learning styles. You can still learn these skills and apply them.
@TedShifrin I'm actually a calculus 1 student. But during the COVID-19 period, I learnt many things in maths which are not even related to my course. Recently I saw the symbol $\oint$ and was greatly attracted to that symbol LOL. And then I started learning complex analysis.
good afternoon. another sweltering 21c day here in albany, ca
my dad was in finance and he had some schaum outline book related to financial calculations (we're talking 60s here). i think my gateway to math was all the $\sum_{n=1}^N$ symbols.
i was stymied by a simple convex question recently, it ended up with a simple answer on MO that took me hours to unwind.
My son is in 2nd grade. His math teacher gave the class a quiz, and one question was this:
If a triangle has 3 sides, and a rectangle has 4 sides,
how many sides does a circle have?
My first reaction was "0" or "undefined". But my son wrote "$\infty$" which I think is a reasonable answ...
Are the following definitions of relative countable compactness equivalent?
$\overline{A}$ is countably compact
Every sequence in $A$ has a convergent subnet
Note that for regular spaces, the analoguous definitions are equivalent for relative compactness. Also, the equivalence always holds if...
