Given that there is an isomorphism $f$ from G to H If $ n \in \mathbb{Z}^{+}$ and $G$ has a n element of order $n$ then $H$ has an element of order $n$.
Hi guys, got a question: "Show that a connected metric space with at least two points contains no isolated points". Can anyone check if this is the easiest way to prove this?
Suppose in connected MS $(X,d)$ there is an isolated point $x \in X$, so that $\exists r > 0$ s.t. $B_r(x) \cap X = \{x\}$.
Then we argue that the set $\{x\}$ and $X \cap \{x\}^c$ is separated:
To prove the first part, i.e. $\overline{A} \cap B = \emptyset$
we see that $\{x\}$ is a single point so it's closed, thus its closure is the same. Then given disjointness the above must be true.
To prove the second part, i.e. $A \cap \overline{B} = \emptyset$, see that $x$ is not an accumulation/limit point(because it is an isolated point). Then there are no sequences in $X \cap \{x\}^c$ that converges to $\{x\}$ (we can't take arbitrary $\epsilon$ because for any $\epsilon$ less than $r$, we cannot provide the definition of convergence to $x$).
Then $x$ is not in the closure of $B$, so the intersection must be empty. Then it contradicts the connectedness of $X$.
I know for sure the trivial subgroups 1,12 will be normal. However, I am not sure how to find the other one(s) if they exist. I don't have Sylows theorem available to me
Today after my presentation at the grand opening I met the president of an engineering company. He really liked my fish feeder design and said he wants me to present it to his engineers once I finish the prototype. He gave me his business card and everything!!
It saved me from getting in trouble too. My English teacher called my mom and said I'm getting an F if I don't turn in my overdue essay. My mom didn't even get mad. But I have to stay up tonight and do it.
It was a lot of work to do a good job. I did a year-long applied math course back 30 years ago. It was a great course, but a ton of work. I did probability right before I retired. That took a good deal of thought/preparation. I did some stuff in grad courses I'd never learned/studied, too.
@TedShifrin I am not sure how to begin. I just write $p_i=(x_{1,i}, \cdots, x_{n,i}$ and try to make these points satisfy some equation. But this seems complicated. How about considering the case where some of the points are collinear?
@AlessandroCodenotti You asked me if the Hawaiian earring is an inverse limit a couple weeks ago and I said no. I think I was wrong, which I realized while writing this answer.
I guess if this is right then it is a counterexample to "$\pi_1$ commutes with inverse limits"
Sure. You get a system of $n-1$ linear equations $A\cdot (x_i-x_1) = 0$, for $i=2,\dots,n$ (for the entries of $A\in\Bbb R^n$). So you expect a $1$-dimensional solution space (which gives $A$ up to scalar multiples). If the points are not in general position, the rank is $<n-1$ and you have infinitely many hyperplanes.
So @mathsresearcher you have two cases, The first case is that the first card is a king of spades, The second case is that the first card is a spades but not a king
The first cases' probability will simply be $\frac{1}{52} * \frac{3}{51}$
Can you tell me the probability for the second case?
Once you have found that you can add and use the formula for P(A|B) "Probability of A given B" to find the answer
but we are trying to find the probability that A happens given that B happens
Thus we must divide out the Probability of B happening
Lets say we have groups $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_5$, $\mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_5$, and $\mathbb{Z}_8 \times \mathbb{Z}_5$
Which of these have an element of order 20
How do I go about answering this without doing it exhaustively and checking manually?
Nvm I think I figured it out I forgot about the theorem which tells us order of elements of direct product is lcm of the order of the individual elements
I'm calculating the fundamental group of a $g$-holed $k$-punctured torus $T_{g,k}$ where $k,g>0$.
I'm going about it with Van Kampen, with one set being a {$4g$-gon with edges identified without a disk} and the other being a {disk with $k$ holes}.
The $4g$-gon with the disk is a bouquet of $4g$ circles, so it has the fundamental group $F_{4g}=\langle a_1,a_2,\dots, a_{4g}\rangle$ (the free group on $4g$ generators).
The torus with a disk missing has fundamental group The disk with $k$ holes has fundamental group $F_{k}=\langle b_1,b_2,\dots, b_{k}\rangle$. The overlap $A$ is an annulus with free group $\mathbb{Z}=\langle c\rangle$.
We can solve the relation for $b_k$ in terms of the other generators, so we can toss out the relation and $b_k$. This leaves us with $\pi_1(T_{g,k})=F_{4g+k-1}$.
Here's a trick: Note that surface of genus $g$ minus a point deformation retracts to the 1-skeleton, the bouquet of $2g$ circles. Then surface of genus $g$ minus $k$ points deformation retracts to bouquet of $2g$ circles wedge a bouqet of $k-1$ circles.
So the space is homotopy equivalent to wedge of $2g+k-1$ circles
@Cbjork By this argument, yes
A wedge of circles has trivial higher homotopy group: it's got a contractible universal cover!
I read the Wikipedia articles for both topology, graph theory (plus topological graph theory). Does topology encompass also graph theory? Or topology is only about studying shapes while graph theory is about relations and the two meet in topological graph theory?
Oh, oh know what, the image I drew above is the same as this one from Wikipedia, except that the top-right strand in Wikipedia's image has been moved to the bottom in mine
@Secret This doesn't make any sense at all. Firstly your indentation is clearly wrong. Secondly at one point you wrote "Q(0)" for no apparent reason at all. Can you please follow the rules strictly?
I am a naturally born rule breaker, no wonder why I found logic so hard to wrap my mind around
Let $X$ be a topological space, $\{f_i\}$ is collection of continuous functions from $X$ to $\Bbb{R}$. Suppose that $f(x) = \inf_{i \in I} f_i(x)$ exists for each $x \in X$. Does it follow that $f$ is continuous?
I'm kind of the same way, Secret, but it's more conscious curiosity on my part. Like, I am interested in axioms of mathematics, solely because I enjoy the idea of adding axioms specifically to see what breaks.
@user193319 What could the collection of continuous functions look like?
OK, so this is the last I'm gonna write about the Perko knot pair on this chat, for a while at least, I promise, but I just wanted to share that I wrote a question on it:
In 1974, a paper titled On the Classification of Knots appeared showing that the knots $10_{161}$ and $10_{162}$ in Dale Rolfsen's knot table were actually the same knot. He included this picture, showing how to deform one into the other:
From that point on, $10_{161}$ and $10_{162}$ became kn...
[Given] Let Q(n) denote "Forall k in N ( k<n implies P(k) )".
[Given] Then Q is a property on N.
[If sub] If Forall n in N ( Forall k in N ( k<n implies P(k) ) implies P(n) ):
[If sub] Forall n in N ( Q(n) implies P(n) ).
[Forall elim] Q(0) implies P(0)
[If sub] If Q(0):
[If sub] Q(0)
[or intro] Q(0) or not Q(0)
[Given intro] Given k in N:
[Implies intro] If Q(0):
[Implies intro] If k < 0:
[Implies intro] If 0 < 0:
Just how on earth I keep going in circles without realising until I get to the end of the line
@guy that responded to me this is not programming, but possibly an infidel has been allowed to edit mathematics Wikipedia articles which would explain why $\sum_{n \operatorname{mod} c}$ has been used, but I don't know what a representative is
I read the Wikipedia articles for both topology, graph theory (plus topological graph theory). Does topology encompass also graph theory? Or topology is only about studying shapes while graph theory is about relations and the two meet in topological graph theory?
In 1974, a paper titled On the Classification of Knots appeared showing that the knots $10_{161}$ and $10_{162}$ in Dale Rolfsen's knot table were actually the same knot. He included this picture, showing how to deform one into the other:
From that point on, $10_{161}$ and $10_{162}$ became kn...
