While looking around stack exchange I noticed this: ∀x∈U:q(x)≡∀x:[x∈U→q(x)] http://math.stackexchange.com/questions/2124099/simple-question-about-universal-quantifier
I have an idea how to solve it so I would like to discuss it with you. Show that $\mathbb{I} \times D^n$ has $\{0\} \times \mathbb{D}^n \cup \mathbb{I} \times S^{n - 1}$ as deformation retraction
So I consider the point $(2,0) \in R \times D^n$
okay ?
I project it onto the cylinder and I get the required homotopy.
so the way I am thinking about this is first given a point $(s,x) \in I \times D^n$ I want to determine the end point for which the ray goes to in $I \times D^n$
with me so far ?
once I have an end point for each (s,x) I can use straight line homotopy to get the required deformation retraction.
Hi, is it true that $\frac{\partial}{\partial q} = \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial}{\partial \dot{q}}$, where $\dot q = \frac{\mathrm{d}q}{\mathrm{d}t}$?
@arctictern Well, intuitively it's 0, since $q$ doesn't depend on $\dot{q}$. But it may be that my intuition may be wrong because I think about Classical Mechanics, where $q$ is a generalized coordinate and $\dot{q}$ is a generalized speed. Is this answer correct?
sometimes we have some multivariable function F(-,-) we want to evaluate at q and q' (and t), in which case to consider a partial derivative of F evaluated at q and q' you have to pretend q and q' are independent
(e.g. lagrangians, as semi said)
@mikeonly okay, so what happens if we apply both sides of your quality to $q$?
@arctictern That's would be a contradiction, since $\frac{\partial}{\partial q} q = 1 \neq 0 = \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial q}{\partial \dot{q}}$, right?
I'm trying to think if there's a thermodynamic application of that identity. It wouldn't shock me, given how much that subject loves partial derivative relations.
@arctictern But is that true that $\dot{q}$ doesn't depend on $q$? It's physically sounds right, but mathematically we can express coordinates in terms of speed.
@mikeonly of course $\dot{q}$ depends on $q$, you just temporarily pretend that $\dot{q}$ and $q$ are different letters of the alphabet when partially differentiating whatever expression you have for $F(t,q,\dot{q})$
We arrange the rational numbers in a line. Then, we choose a position to apply a cut in the middle, so that everything left of it is a set and everything right of it is another set. Then, there are $2^{\Bbb Q}$ possible cuttings.
@DHMO Well, given a sequence of $N$ elements, you can place a cut in $N$ different spots. But that doesn't give you all partitioning of $N$ elements, since by rearranging them you get different partitioning. Say, for a given line and all possible cuts there are no such sets, where the first and the last elements are together, if cut is not at the very beginning, or at the very end.
I still think that by going through $\mathbb{Q}$ and choosing whether an elements belongs to $A$ or to $B$ (where $A$ and $B$ partition the rationals) you get $2^\mathbb{Q}$.
@DHMO I suspect bigger. Note that you are talking about the set of dedekind cuts, which is not the set of bounded sequences of rationals. (And Ted amended his comment to now mean set of bounded sequences of rationals, modulo an equivalence relation.)
@arctictern I think each element in the set can be represented by an element of $\Bbb R \times \Bbb R$, namely the infimum and the supermum, so it has cardinality $\Bbb R$, but I might as well be using my intuition badly
We arrange the rational numbers in a line. Then, we choose a position to apply a cut in the middle, so that everything left of it is a set and everything right of it is another set. Then, there are $2^{\Bbb Q}$ possible cuttings.
@Semiclassic: We're intersecting two divisors (= sums of formal curves) in a surface. The curve $C$ with itself, and then $C$ with the canonical divisor (for which we have a formula, depending on the surface — in the case of $\Bbb P^2$, it's $-3$ times a line).
@Semiclassic: Hmm, Georges does not make mistakes. So the two formulas are at odds. I'll have to check that the formula your first post starts with is consistent with the one Georges starts with. I'll try to get to this on Friday.
In mathematics, a square-integrable function, also called a quadratically integrable function, is a real- or complex-valued measurable function for which the integral of the square of the absolute value is finite. Thus, if
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hello, i'm new to graph theory, i'm learning basic things, eular circuit, path ,hamiltonian path, isomorphic graphs , can anyone please suggest me which books should i follow for practice problems with solutions
Suppose $f:2 \to 1$ is injective. From the definition of $1$, $x \in 1 \equiv x = 0$. From the definition of injective, $f(0) \ne f(1)$. However, $f(0),f(1) \in 1$. Therefore, $f(0) = 0$ and $f(1) = 0$. Contradiction.
Suppose $f:1\to0$ is injective. From the definition of $\to$, $f(0) \in 0$. However, $\not\exists x:x\in0$. Contradiction.
@DHMO What if you assume that $k^+$ injects into $k$, but $k^{++}$ doesn't inject into $k^+$?
Or by induction, from your previous argument you want to show that since there is no injection from $k^+$ to $k$, then $k^{++}$ does not inject into $k^+$ either.
Well, let $k^{++} - x$ be such a set after you removed a point from this set. Since you assume that $\exists f$ injection, $f: k^{++} - x \rightarrow k^{+} - f(x)$ is still an injection.
@DHMO Now, take a bijection from $k^{++} - x$ to $k^+$ and do a similar thing with $k^{+} - f(x)$. Now, by induction hypothesis, there is no injections from $k^{+}$ to $k$, but composition of bijections and $f$ gives you an injection. Therefore, contradiction.
@DHMO What if you enumerate all elements in $k^+ - f(x) =k^+ -1$ and $k$ by order? Assume there is such $k_0 \in k$, which is not in $k^1 - 1$. Then for each $k < k_0$ you can map bijectively from $k^+ - 1$ to $k$. Map all $k \geq k_0$ to $k + 1$, which will also give you a bijection. Will the second map work from the definitions of those sets $k$?
If there is no such $k_0$ in $k$, then you can map to $k^+ - 1$ just by identity.
I had to sanity check myself super hard earlier on whether there can exist a function $f : A \rightarrow \emptyset$ if $A$ is nonempty. (No.)
Mostly because I was like, "I know $\emptyset$ is initial in Set, but why isn't it final like the singletons, with the unique morphism as the empty function?" (Because.)
$\forall a: a \in A \implies f(a) \in \varnothing$. However, $\nexists x: x \in \varnothing$. Therefore, $\nexists a: a \in A$. Therefore $a = \varnothing$.
@PVAL How difficult is it to write down a careful proof of existence of proper Morse functions? ie could you or I do it? I don't really want to think very hard about it.
It seems to me I can do it if I know the existence of a proper function to [0,\infty) but that it would be a pain.
OK yeah I have a proof. But I don't know if it's something I would assign to students.
I want to simply the expression $(A\cup B)\cap (A\cup B')\cap (A'\cup B)$. Using the formulas $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$ and $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$ I found thatthe expression is equivalent to $A\cup B$. I this correct?
As I procrastinate studying for my Maths Exams, I want to know what are some cool examples of where math counters intuition.
My first and favorite experience of this is Gabriel's Horn that you see in intro Calc course, where the figure has finite volume but infinite surface area (I later learned...
