> In difficult situations we often assume that someone else will do something. But if we all assume someone else will step in then nothing happens. We can all try and act on our values in a way that is safe and appropriate – not leave it to someone else.
My book says that a map between Stone spaces $X$ and $Y$ is continuous if for each $U\in\operatorname{Clop}(Y)$ we have $f^{-1}(U)\in\operatorname{Clop}(X)$. I know that a Stone space can contain open sets that are not closed, so apparently we don't need to explicitly check that $f^{-1}(U)$ is open for open $U$, as long as we checken the clopen condition.
a Stone space is a compact, totally disconnected space. it's equivalent to saying that it is compact, Hausdorff and has a basis of clopens
I wanted to show that if clopen->clopen, then we have a continuous map
Is there a name for the cartesian product of two simplices? For example, the space of (row-)stochastic matrices is a cartesian product of simplices. For a 2x2 matrix, that space looks like a square to me. Not sure about 3x3, though.
Or, please provide a counterexample to this statement:
Let $E$ be a dense subset of a metric space $X$, and let $f$ be a uniformly continuity function from $E$ to a (not necessarily complete) metric space $Y$. Then $f$ has a continuous extension from $E$ to $X$.
@TobiasKildetoft Is this function uniformly continuous? I don't think so, because, if your function is from $[-1,1]-\{0\}\to \Bbb R$ then it violates the assertion that 'uniform continuous function from dense domain (dense in $X$) can be extended to $X$ continuously, where codomain is complete metric space'.
I can give a formal statement if you wish.@TobiasKildetoft
I have following thoughts on this suggested counterexample (its in my book, as well as mentioned by you a few days ago): Take $X$ to be real numbers, $Y$ and $E$ be the rational numbers and let $f:E\to Y$ be given by $f(x)=x$. There is no possible extension of $f$ to a mapping from $X$ into $Y$.
Suppose there is an extension of $f$, say $g$ to a mapping from $X$ into $Y$ which is continuous, then let $p_n$ be sequence of rational numbers that converges to $\sqrt 2$. Then since $g$ continuous, $g(p_n)\to g(p)$ but since $g(p_n)=p_n$, this means that $g(p_n)\to\sqrt 2$, but $\sqrt 2$ not in co-domain.
Hello, are the variables a from the second image and x from the definitions listed at en.wikipedia.org/wiki/Hyperbolic_function the same variable? In other words, for sinh x = (e^x - e^-x)/2 and cosh x = (e^x + e^-x)/2 is the area of the red region in the image x/2?
@KasmirKhaan I came by because of a flag; ordinarily this should be the job of room owners. Consider adding more if you feel this room is undermoderated.
for my part I don't actually mind the messages. I find them irrelevant to my own interests, usually, but not offensive. (by contrast, I find the random questions/appeals for help to be a bit exhausting at times)
That said, if you want to have a place to ramble it's just as easy to make a room of your own
(I did that recently for some saddle-point stuff I wanted to get out of my head)
There was this problem on my algebra test today. "Let $n \in \mathbb{N}$ and let $(G, \cdot)$ be a cyclic group such that $|G| = n$. Let $(H, \star)$ be a group and for each $y \in H$ we have that $gcd(n, o(y)) = 1$. Show that there exists exactly one group homomorphism from $G$ to $H$."
I think $H$ was finite in the statement of the problem
@Tobias How could you go about solving it? All I can say at the moment is that since $(G, \cdot)$ is cyclic, we'll have $G = \langle a \rangle$ for some $a \in G$, and $o(a) = n$.
I'm thinking I'm gonna have to construct a homomorphism from $G$ to $H$, then suppose another homomorphism exists and arrive at a contradiction somehow
Ahh $f[G = \langle x \rangle] = \langle f(x) \rangle$. Since $o(x) = n$, $o(f(x)) | o(x)$
Ignore the above message
Suppose $f$ is a homomorphism from $G$ to $H$. Since $G = \langle x \rangle$ we have $f[G] = f[\langle x \rangle] = \langle f(x) \rangle$. Now since $o(x) = n$ is finite, and $f$ is a homomorphism, $o(f(x))$ divides $o(x)$.
Suppose $x$ is a binary string of length $n - k$, where $n, k$ are integers and $n > k$. An 'extension' of $x$ is a string of length $n$ which is formed from $x$ by inserting exactly $$k$$ total new 0s and 1s at any position in the string $x$. How many distinct such 'extensions' are there from $x$? Does it depend on more than the length of the string $x$ (i.e., the value of $k$)?
Based on numerical evidence the answer is 'it does not depend on the actual string x'.And the number of distinct extensions is $\sum_0^k \binom{n}{k}$
So pick any $g \in G$, then note that $g = x^m$ for some $m \in \{0, 1, ..., n\}$, then observe that $o(f(g)) = o(f(x^m)) = o(f(x)^m) = o((e_H)^m) = o(e_H) = 1$. This again only occurs if and only if $f(g) = e_H$, hence $f(g) = e_H$ for every $g \in G$. Hence $f : G \to H$ is defined by $f(g) = e_H$ for every $g \in G$.
Since we've picked an arbitrary homomorphism from $G$ to $H$ and shown that it equals a specific homomorphism, there can only be one such homomorphism and we're done
For a, b, c ∈ Z, prove a | b if and only if ca | cb. My attempt: suppose ca = cb. Since c | a and c | b then a | b by the theorem of the GCD--Divisibility equivalence.
What is the smallest positive integer n for which 147x+105y = n = 606u+909v for some x, y, u, v ∈ Z? For this n, give values of x, y, u, v that satisfy the requirements. My attempt: I think n is 3, since 3 is common between all of them and divides them all. Then it requires some messy Euclid's algorithm
@DarkVampiricAbstractArtist: Probably easiest to do it a few at a time. Find the gcd of the first two, find the gcd of the last two, then the gcd of the two gcds.
Bezout's Identity is what I call the Euclidean algorithm, by the way.
