I appreciate the suggested immortality, but I've spent most of today working out how to prove something I learned a while ago, so that I can prove something that's probably going to turn out to be trivial.
@ForeverMozart Usually the approach is to either stare at it until I've broken down and started crying, at which I go home for the night, or (if I know something similar has been done somewhere else) to look at other things that deal with similar problems.
I am very envious of this other graduate student. He is very clever at defining nice sets, proving they have various properties, and then using them to prove his results.
@AkivaWeinberger I may have no idea what I'm talking about. The subset graph for N is just one of my favorite Axiom of Choice graphs that doesn't require the real numbers.
> Every finite connected graph has a spanning tree. However, for infinite connected graphs, the existence of spanning trees is equivalent to the axiom of choice. An infinite graph is connected if each pair of its vertices forms the pair of endpoints of a finite path. As with finite graphs, a tree is a connected graph with no finite cycles, and a spanning tree can be defined either as a maximal acyclic set of edges or as a tree that contains every vertex.
I try to use a piece of advice from Feynman. He said to no try too hard for a particular result, but instead to simply try to find out more about the object your are studying.
So if I am trying to prove something about a particular type of space, and it's too difficult, I try to prove some easier things about it.
luckily in topology there are lots of different properties to think about
There was a "horror story" I heard of, once: A student was preparing his PhD thesis on some sort of weird object (I forget what it was, I think it was some weird variant on metric spaces). …
In every topological space the empty set and the one-point sets are connected; in a totally disconnected space these are the only connected subsets (quoting en.wikipedia.org/wiki/Totally_disconnected_space)
I try to forget everything besides topology, algebra, and graph theory, and maybe some discrete geometry. Those are the most beautiful areas and I don't have time to waste on anything else.
I am not passionate about the fact that your category has an initial object. It's not even really clear to me why I should care. But the game we were playing wasn't the existence of the empty space, but about the empty quantifier over 1-point sets.
There are a lot of other games I would rather play, like smash bros.
I just saw some disagreement about whether or not the empty set is a topological space. The question I asked was just one I had previously come across where I genuinely had to consider this.
People tend to ignore the distinction between the topological space and the underlying set. Just like how you should ignore any sentence beginning with "technically" :P
Let $ f: D \rightarrow R$ with $x_{0}$ as an accumulation point of D. Prove that f has a limit at $x_{0}$ if for each $ \epsilon >0$ there is a neighborhood Q of $x_{0}$ such that, for any $x,y \in Q \cap D, x \neq x_{0}, y \neq y_{0},$ we have $\mid f(x) - f(y) \mid < \epsilon$. \\
By the definition of a limit for the function and Theorem 2.3, we have $\epsilon >0$ and f has a limit at $x_{0}$, so there exists a $\delta$ such that for any $x \in ( x_{0} - \delta, x_{0} + \delta)$, we have $ \mid f(x) - L \mid < \frac{\epsilon}{2}$.\\
I'll get theorem 2.3 in a sec. it's about accumulation point and the neighborhood
2.3 THEOREM Let $ f : D \rightarrow R$, with $x_{o}$ an accumulation point of D. If f has a limit at $x_{o}$, then there is a neighborhood Q of $x_{o}$ and a real number M such that for all $ x \in Q \cap D \mid f(x) \mid \leq M$
If $f_{\alpha}:X_{\alpha}\rightarrow Y_{\alpha}$ is bijective for each ${\alpha}$ in index set $A$ then $f:X\rightarrow Y$ defined as $f(x)=\prod_{\alpha\in A}f_{\alpha}(x_{\alpha})$ is bijection where $X=\prod_{\alpha\in A}X_{\alpha}$ and $Y=\prod_{\alpha\in A}Y_{\alpha}$
Does anybody have an approachable explanation of differential equations? My textbook (Stewart; Early Transcendentals) just throws undecipherable examples of what they look like at me.
I have looked at a large number of sources, including those on the main site.
What I'm really looking for isn't an explanation of what they are, but how to approach them and why those approaches work. I'm working with ordinary DEs with order 1. I'm using separation of variables.
Bit of a soft question coming from a noob... Suppose $h$ is irrational. What can be said of the set of all rational multiples of $h$? Are there any peculiar differences given whether $h$ is algebraic or otherwise?
My question is mainly motivated by the set $\{q\pi\in\mathbb{R}:q\in\mathbb{Q}\}$ and its size and properties compared to the continuum.
I was told to give an example of a monoid . My example was $\frac{\Bbb{Z}}{2\Bbb{Z}}$ under multiplication of residue classes defined by $\overline{a}\overline{b}= \overline{ab}$
Hey guys, I'm going through Clement's theorem on Twin Primes, and at one point he says 2[(-1)(-2)(n-1)!+2]+n is congruent to 2[(n+1)!+2]+n mod(n+2), but I can't see how.
In a dihedral group is rotation taken clockwise or anticlockwise? Or does it really matter? Because If we take one rotation clockwise it is the same as taking three rotation anticlockwise
Let $\phi: \mathbb{C}^{\times} \to \mathbb{R^{\times}}$ be defined by $\phi(x+iy) = x^2+y^2$. Prove that $\phi$ is a homomorphism and describe the kernel and image of $\phi.$
I think $|(x+iy)^2| = |x+iy||x+iy| = x^2+y^2, $ so $\phi$ is a homomorphism. $\text{ker}(\phi) = \left\{x,y \in \mathbb{R}\setminus \left\{0\right\}: x+iy = 1\right\} = \left\{r \in \mathbb{R}\setminus \left\{0\right\}: re^{i\theta} =1 \right\}$? Is this right? And $\text{im}(\phi) = \left\{x+iy: x,y \in \mathbb{R} \setminus\left\{0\right\}\right\}$?
I think confused my definitions, it should be $\text{ker}(\phi) = \left\{x+iy \in \mathbb{C}: x^2+y^2 = 1\right\} = \left\{z \in \mathbb{C}: |z| = 1\right\}$ which is the unit circle.
@barrycarter @TheKindCat $x^2=y^2$ for $x\ne y$ in a group does not imply that there is an element of order two. It also does not imply that $xy^{-1}=yx^{-1}$.
mmkay. i don't notice anything particularly special about that, though, aside from that set being the same as $2\mathbb{Z}_{10}\cong \mathbb{Z}_5$. (and that doesn't really depend on $a\in\mathbb{Z}_{15}$ aside from $\mathbb{Z}_{15}$ containing $\{0,1,2,3,4,5\})$
@Semiclassical Perhaps I then got something wrong: I was calculating the image of the homomorphism $\phi: \mathbb{Z}_{15} \to \mathbb{Z}_{10}$ defined by $\phi([a]_{15}) = [2a]_{10}$.
