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4

Yes, you're doing it right, however you have to use the fact that words are finite in order to prove that this is antisymmetric. Consider the infinite case, the word $abababab\ldots$ then it is a substring of the word $babababa\ldots$ and vice versa, but they are not equal. So it is a partial order. The other two do not hold, but I'll leave it to you to ...


3

In your case, roughly speaking you could say "one can get $f(x)$ as close as imaginable to $L$, by setting $x$ close enough to $a$". About the second statement, it is not right, since there may be values of $x$ for which $f(x)=L$ but $x\neq a$, or $x$ is "very far" from $a$. For example, think of the constant function $f(x)=L$. Then, $$ f(0)=L = \lim_{x\to ...


2

Depends on the construction you make of $\mathbb{R}$. Assuming that you already have built the operators $+,\cdot$, a simple way to notice it would be to make a subset $\mathcal{P}\subset\mathbb{R}$ which represent the positive numbers, then $a\leq b \iff b+-a\in \mathcal{P}$, where $-a$ is the additive opposite of $a$ such that $a+-a = 0$.


2

It all comes down to how we're setting the reals up. If we're using Dedekind cuts, then yep, that's how to do it! If we're using equivalence classes of Cauchy sequences, then we have to say something messier: $a<b$ if, whenever we pick representative sequences $\alpha\in a$ and $\beta\in b$, there is some $n$ such that for all $m>n$, ...


2

If you define the reals with Dedekind cuts, yes: $$\forall a,b\in\mathbb R(a\leq b\iff a\subseteq b)$$ $$\forall a,b\in\mathbb R(a<b\iff a\subsetneq b)$$ It's a little harder to define ordering on the reals if you use Cauchy sequences. (This is because Dedekind cuts are specifically designed to "complete" the ordering of the rationals, while Cauchy ...


2

The way i think about this definition(please correct me if I'm wrong) is that for any given range ϵ around L, we can find some range δ around x such that the function evaluated at those points will be within ϵ of L Yes. Rephrasing: given an error $\epsilon$ around $L$, you can find a safety margin $\delta$ around $a$ such that if $x$ is withing that ...


2

The second isn't even an order, since an order has to be antisymmetric...i.e if $xRy$ and $yRx$, then $x=y$ for any order. In the case of 2, you have $(0,1)R(1,0)$ and $(1,0)R(0,1)$ but $(1,0)\ne (0,1)$


1

You have many questions, I'll try to adress them all. A binary relation, as you read is just some set $R$ which is a subset of the cartesian product of two sets $A$ and $B$, that is $R \subseteq A\times B$. An example may ilustrate this: Let $A=\{\dots,-4,-2,0,2,4,\dots \}$ (the set of even numbers), $B=\{1,3,5\}$. Then a relation $R_1$ could be ...


1

Clearly, it suffices to define lines in a order-theoretically way, as $X$ is affine if and only if it contains (it is bigger than) all the lines through its points. All sets are supposed convex. The order relation is "$X$ is smaller thatn $Y$ iff $X\subset Y$" Definition: X is polygonal if $\exists$ a convex $Y\supset X$ such that $Y\setminus X$ is ...


1

Given a $C^{\infty}$ manifold $M$, a vector field $X$ is a derivation on the algebra $C^{\infty}(M)$ of $C^{\infty}$ real functions on $M$. That means: $X$ is a map $X: C^{\infty}(M) \rightarrow C^{\infty}(M)$ that respects the following properties: (i) $X$ is linear: $X(\alpha f + \beta g)=\alpha Xf+\beta X g; \quad \alpha, \beta \in \mathbb{R}, \quad ...


1

Let $X = a\partial_x +b \partial_y$. Then, $$ [\partial_x, X]f = \partial_x(X(f))-X(\partial_x f)$$ But, $X(f) = a\partial_xf +b \partial_yf$ and likewise for the second term, hence: $$ [\partial_x, X]f = \partial_x(a\partial_xf +b \partial_yf)-a\partial_{xx}f +b \partial_{xy}f$$ Of course, $a,b$ are functions thus there are product rules to consider in the ...


1

A partial order, $\leq$, satisfies the following properties for all $a,b,c$: Reflexivity: $a\leq a$ Anti-symmetry: If $a\leq b$ and $b\leq a$ then $a = b$ Transitivity: If $a\leq b$ and $b\leq c$ then $a\leq c$ A partial order however does not have to have defined comparisons between all elements. A total order must satisfy all of the same properties ...


1

You're mostly right. First, $\epsilon$ can get arbitrarily close to $0$, not $L$. In the definition we write $$ \left | f(x) - L \right | < \epsilon $$ So $L$ does not have anything to do with the value of $\epsilon$ directly. Also, there is no requirement for $\delta$ to be arbitrarily small. The best way to understand $\delta$ is to consider it as a ...


1

In Dedekind's construction $\alpha < \beta$ is defined to mean $\alpha$ is a proper subset of $\beta$. What you say is correct.


1

A stochastic matrix is, by definition, any matrix for which each row, column or both, sums to $1$. Mathematics does not concern itself with what mathematical objects "represent". It only cares about what their properties are. And what you described is a stochastic matrix. In fact, every stochastic matrix will be a transition matrix for some Markov chain, ...


1

The question means the following: Let $S$ be some set. Let $P$ be the power set of $S$; that is, let $P=\mathcal{P}(S)$. Thus, elements of $P$ are themselves subsets of $S$. Given any two $X,Y\in P$, we can define their set difference $Y-X\in P$ to be this subset of $S$: $$Y-X=\{s\in S: s\in Y\text{ and }s\notin X\}$$ Define a relation ...


1

The only alternative I can think of would be to consider everything as a non-unital ring, a.k.a. "rng" (Wikipedia link). Then the map $\phi:R\to\mathrm{M}_2(R)$ is a perfectly fine rng homomorphism, and since the rng $R$ happens to have an identity, so will the image $\phi(R)\subset\mathrm{M}_2(R)$. If you want to stick with rings instead of rngs, I would ...


1

Two points: One, Galois closure is a relative concept, that is not defined for a filed, but for a given extension of foields. Second, it is not something maximal. To the contrary it is something minimal. Given an extension of fields $F\subset E$ if it is not Galois, then the smallest extension of $F$ that containing $E$ and that is a Galois extn of $F$ is ...



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