How to show, that for every continuous $f:  X\rightarrow X$ there exists $x \in X$, such that $f(x) = x$, where X is a real projective plane $\mathbb{R}P^2$.  In other words: every continuous map of RPP to itself has a fixed point.   EDIT  Probably it's easier to proof, that existence of fixed point in map from $S^2$ to itself implies needed fact.
Let $f:S^{2n} \rightarrow S^{2n}$ be a continuous map. Show that          there exists $x \in S^{2n}$, such that $f(x) =x$ or $f(x) = -x$;   any continous map $g: \mathbb R P^{2n} \rightarrow \mathbb RP^{2n}$ has a fixed point;      I think 1 implies 2. So, how can I find an $x$ in 1?  Thanks for the help.
I don't know if this is appropriate for math.stackexchange, or whether philosophy.stackexchange would have been a better bet, but I'll post it here because the content is somewhat technical.  In ZFC, we can prove that the (second-order) Peano axioms have a model, call it $\mathbb{N} = (N,0,S)$. Furthermore, $\mathbb{N}$ is unique up to isomorphism. Thus, it would seem that we've pinned down $\mathbb{N}$.  However, if ZFC is consistent, then it has some very peculiar models. In particular, it has a model $(V_0,\in_0)$ whose "native" $\mathbb{N}$, let call it $\mathbb{N}_0$, is a model for the sentence "ZFC is consistent," and another model $(V_1,\in_1)$ whose "native" $\mathbb{N}$, lets call it $\mathbb{N}_1$, is a model for the sentence "ZFC is inconsistent."  But in reality, in other words, for the "real" $\mathbb{N}$, only one of these can be sentences can be true. Furthermore, the objects $\mathbb{N}_0$ and $\mathbb{N}_1$ aren't isomorphic, since the set of sentences they satisfy are different.  So it seems to me that $\mathbb{N}$ cannot truly be pinned down. Furthermore, its categoricity is, in some sense, illusory. Is this correct, or am I missing something? I am studying homology groups and I am looking to try and develop, if possible, a little more intuition about what they actually mean. I've only been studying homology for a short while, so if possible I would prefer it if this could be kept relatively simple, but I imagine it is entirely possible there is no real answer to my query anyway.   As I said above, I want to gain a little deeper understanding of what the n-th homology group actually means: I can happily calculate away using Mayer-Vietoris but it doesn't really give me a great deal of intuition about what the n-th homology group actually means. For example, with homotopy groups, the fundamental group is in some sense a description of how loops behave on the object in question, and it is obvious to me why that is what it is for say, the torus or the circle. However, I have no idea what, if anything, I am actually saying about a triangulable object when I talk about it having 0-th homology group this or 1st homology group that.  The best I have been able to find online or in my limited book selection is the brief description "intuitively, the zeroth homology group counts how many disjoint pieces make up the shape and gives that many copies of $\Bbb Z$, while the other homology groups count different types of holes". What 'different types of holes' are there, roughly speaking? I appreciate that it may often be completely non-obvious what the low-order homology groups are for some complicated construction, but perhaps in simpler examples it might be more explicable. Are there (simple) cases where I could say, just from looking something like e.g. the torus, what its zero-th or first or second etc. homology group was based on the nature of the object? I guess in the zero-th case it is, as my source () above says, related to the number of disjoint pieces. Can we delve deeper than this for the other homology groups?   Any book/website suggestions would be welcomed (preferably websites as I am nowhere near a library!) - I have Hatcher but not a great deal else, and I haven't gleaned as much as I wish to from that alone. Of course I know that there is a great deal we don't know about homology groups even today, so I don't expect some magical all-encompassing answer, but any thoughts you could provide would be appreciated. I hope this question is appropriate for SE Mathematics, apologies if not! -M Is there a case insensitive version of the  jQuery selector or should I do the work manually by looping over all elements and comparing their .text() to my string? Proof of infinitely many primes, clarification I found this , but I am not sure if this is the same issue.  On , I did some reformatting of the indentation in the code block to make it a bit more readable and it was  for peer review. A few minutes later, it appears the OP made an edit that added some additional code. It looks like my suggested edit was applied when the OP saved his/her edit, and it looks like he got the "credit" for the edit.  I'm not really concerned about the 2 rep points; I'm more curious why this happened and if this is a bug (or limitation) in the editing feature, or if this is known/intended behavior?  Note that this is not a suggested edit that's overridden by a concurrent edit: the recorded edit starts from the post-suggested edit state (and in fact can be identical to it, as in  ). Some examples include:     We fear the damned.   He honored our fallen.   This is a given.   You are the chosen.   The lost were among us.   They obey the venerated.   My beloved kissed me. (TIL “beloved” is a verb thanks to RegDwight АΑA)   Is there a word to describe the past participle used as a noun? I’m hesitant to refer to it as a gerund, since it doesn’t end in -ing. Another possibility is adjectival noun, if you originally treat the word as a participle (e.g. “the damned man”) that has given up its modified noun.  Also, why does English tend to preface such nouns with the? Can I safely charge my laptop with a non-standard, third-party charger? I have a Toshiba Satellite laptop. My charger has stopped working.  I have access to a Lenovo charger. Can I use this charger on my laptop? Property of the conjugate transpose matrix with inner product Object has no method 'live' - jQuery How can one find the zeroes of $f(x)=ae^{bx}+cx+d$? A certain physics problem I have been working on has turned into a math problem. Particularly, I want to find the solutions of some equation of the form   $$f(x)=ae^{bx}+cx+d = 0$$  where $a, b, c,$ and $d$ are constant, real numbers that come from the physics problem and $x$ will be a real number. I do not know how to find the solutions of an equation of this form. In the special case where $c=0$, the solution is simply  $$x = \frac{\ln(\frac{-d}{a})}{b}$$  but for other values of $c$, I am stumped. If there is no exact form for the roots, is there a relatively simple way to estimate the roots? Where can I find information about upcoming special issues in a field of study? Reverse / invert a dictionary mapping Given a dictionary like so:  my_map = {'a': 1, 'b': 2}   How can one invert this map to get:  inv_map = {1: 'a', 2: 'b'} the largest integer $n$ for which $n+5$ divivides $n^5+5$? When we divide $n^5+5$ by $n+5$, we get a remainder $-620$, i.e., $n^5+5=K(n+5)-620$ , now how to proceed further? Is it proper to use "the" before the name of a government organization? Does $f(x)$ is continuous and $f = 0$ a.e. imply $f=0$ everywhere? I wanna prove that   "if $f: \mathbb{R}^n \to \mathbb{R}$ is continuous and satisfies $f=0$ almost everywhere (in the sense of Lebesgue measure), then, $f=0$ everywhere."  I am confident that the statement is true, but stuck with the proof. Also, is the statement true if the domain $\mathbb{R}^n$ is restricted to $\Omega \subseteq \mathbb{R}^n$ that contains a neighborhood of the origin "$0$"? How come Mortarboard is bronze and Enthusiast is silver? :   Silver  Enthusiast    Visited the site each day for 30 consecutive days   56.9k awarded Bronze  Mortarboard   Earned at least 200 reputation in a single day      15.4k awarded    It would seem that earning Enthusiast requires almost no effort while Mortarboard is highly non-trivial. This appears to be reflected in the award count (almost 4:1 in favor of enthusiasts).  So, why is the first one silver and the second one bronze?  (A similar question can be asked about Epic vs Fanatic and probably many other). I need to prove that $$\sum_{n=1}^{\infty} {\sin{100n}} \; \text{diverges}$$ I think the best way to do it is to show that $\lim_{n\to \infty}{\sin{100n}}\not=0$. But how do I prove it? Can you flank with a reach weapon? The Variant: Flanking rules in the DMG (page 251) require adjacency, but is the intent that the Enemy has combatants on opposite sides and thus someone with Reach (and using the reach) could still flank even if not directly adjacent?  Looking for both RAW as well as Intent of the flanking rule. How would you evaluate $\liminf\limits_{n\to\infty} \ n \,|\mathopen{}\sin n|$ How would you evaluate the limit inferior of the sequence $n\,|\mathopen{}\sin n|$? That is, $$\liminf\limits_{n\to\infty} \ n \,|\mathopen{}\sin n|$$    Edit. Let $\mu$ be the irrationality measure of $\pi$. Since $\mu$ is not known, I will split the question:   Assuming $\mu &gt; 2$, what is $\liminf\limits_{n\to\infty} \ n\,|\mathopen{}\sin n|$? Assuming $\mu = 2$, what is $\liminf\limits_{n\to\infty} \ n\,|\mathopen{}\sin n|$?     I kind of like its graph... Calculate limit with L'Hospital's rule I want to calculate the limit $\displaystyle{\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}}$.   I have done the following:    It holds that $\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}=\frac{0}{0}$.   So, we can use L'Hospital's rule:  \begin{align*}\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}&amp;=\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x} \\ &amp;=\lim_{x\rightarrow 0}\frac{\left (x^2\cos \left (\frac{1}{x}\right )\right )'}{\left (\sin x\right )'} =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+x^2\cdot \left (-\sin \left (\frac{1}{x}\right )\right )\cdot \left (\frac{1}{x}\right )'}{\cos x} \\ &amp;=\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )-x^2\cdot \sin \left (\frac{1}{x}\right )\cdot \left (-\frac{1}{x^2}\right )}{\cos x} \\ &amp; =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )}{\cos x}=\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )\right ) \\ &amp; =\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )+\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )\end{align*}   We calculate the two limits separately   $\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )$ :    \begin{equation*}\left |\cos \left (\frac{1}{x}\right )\right |\leq 1 \Rightarrow -1\leq \cos \left (\frac{1}{x}\right )\leq 1  \Rightarrow -2x\leq 2x\cdot \cos \left (\frac{1}{x}\right )\leq  2x\end{equation*} we consider the limit $x\rightarrow 0$ and we get \begin{equation*}\lim_{x\rightarrow 0} \left (2x\cdot \cos \left (\frac{1}{x}\right ) \right )=0\end{equation*}    How can we calculate the limit $\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$ ?