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16

The answer is both yes and no, so this will take a bit of elaboration. There are two ways to define a ring. One of them require the existence of a $1$, the other does not. Let's start with the one that does. In this case, a subring is required to contain the $1$ from the larger ring, and hence no proper ideal can be a subring (as any ideal containing $1$ ...


6

In general, an ideal is a ring without unity - i.e. without a multiplicative identity - even if the ring it is an ideal of has unity. For example $\mathbb{2Z} \subset \mathbb{Z}$ is an ideal but $\mathbb{2Z}$ is not a ring with unity. So if you require your rings to have unity - and a lot of the time one does - then an ideal is in general not a ring. This is ...


5

In an elementary context, $\Bbb W$ means the set of whole numbers. Some books have it as $$\Bbb W=\{0,1,2,\ldots\}$$ while others have it as $$\Bbb W=\{1,2,\ldots\}$$ Because of the ambiguity, I recommend that you avoid the use of $\Bbb W$. For the second meaning use $\Bbb Z^+$. There still is no perfect abbreviation for the first. Either meaning is also ...


3

Just to provide a more or less authoritative reference as to what $\mathbb{W}$ denotes, the following is from page 2 of the book A Transition to Abstract Mathematics by Randall Maddox: As you can see, $\mathbb{W}$ denotes the set of whole numbers, but this notation is often avoided in favor of $\mathbb{N}$, and even $\mathbb{N}$ itself is often clarified ...


3

The limit is $a$, not $\varepsilon$. The point is that one can make $a_n$ as close to $a$ as desired by making $n$ big enough. The absolute value $|a_n-a|$ is the distance between $a_n$ and $a$. How big is big enough depends on how close you want to make $a_n$ to $a$. So $\varepsilon$ is how close you want to make $a_n$ to $a$, and $N$ is how big you ...


2

I’ve not dealt much with the concept, but an example that I find helpful is $\Bbb R^2$ with the product partial order, $\langle x_0,y_0\rangle\le\langle x_1,y_1\rangle$ iff $x_0\le x_1$ and $y_0\le y_1$. Then it’s not hard to check that $\langle x_0,y_0\rangle\prec\langle x_1,y_1\rangle$ iff $x_0<x_1$ and $y_0<y_1$. Consider, for instance, the set ...


2

To get a broader context for the well above relation you can consult any introductory text on domain theory. However, the context of continuity spaces is a bit different, and I prefer to have the intuition for Flagg's value quantales come directly from their intended role. So, the way I think about the well above relation is that it solves some nasty ...


2

When a group acts on a graph (in the usual sense), there is more structure than just a permutation of the vertices. Specifically, it not only maps vertices to vertices, but preserves the property that if there is an edge $(a,b)$ in the graph, then there will be an edge $(g(a), g(b))$ as well. That is, it permutes both the vertices and edges in a compatible ...


2

Your relation $R$ is already transitive, so it is its own transitive closure. It appears that what you’re misunderstanding is the notion of transitivity. A relation $R$ on a set $A$ is transitive if it satisfies the following condition: if $\langle a,b\rangle\in R$ and $\langle b,c\rangle\in R$, then $\langle a,c\rangle\in R$. It says absolutely ...


1

The transitive closure of a binary relation $R$ is the intersection of all transitive binary relations that extend $R$. To say that $S$ extends $R$ means that for all $x,y$ in the domain, if $aRb$ and $aSb$. The intersection $T$ of a set of binary relations is defined by saying that for all $x,y$ in the domain, $xTy$ if and only if for every binary ...


1

The natural numbers, $\mathbb N$, are sometimes called the whole numbers, $\mathbb W$. It's ambiguous, because $-1$ is also a whole number, since it has no fractional parts.


1

There are two possible meanings: For any pullback square as below, $$\require{AMScd} \begin{CD} X' @>>> X \\ @VVV @VVV \\ Y' @>>> Y \end{CD}$$ if $X \to Y$ is a monomorphism, then $X' \to Y'$ is also a monomorphism. For any commutative diagram of the form below, $$\begin{CD} X' @>>> X \\ @VVV @VVV \\ Y' @>>> Y \\ @VVV ...


1

In linear algebra a set can generate the space in the sense of linear combinations but if the set is linearly independent then it is a basis. For example in the plane vectors (0,1) and (1,0) constitute a basis because they generate all of plane and are independent while if you consider the set {(1,0),(0,1), v} where v is for example (1,1) then this set ...


1

Note that $X_0\times X_1=\{(x_0,x_1)\mid x_i\in X_i\}$. The product $\prod_{i<2}X_i=\{f\colon 2\to\bigcup\{X_0,X_1\}\mid f(i)\in X_i\}$, it is not a function, it is a set of functions. There is a natural bijection between the two sets. Namely, $(x_0,x_1)\mapsto\{(0,x_0),(1,x_1)\}$. You seem to have confused a typical element of $\prod_{i<2}X_i$, which ...


1

I cannot in good faith answer question 1, because I didn't see enough examples of that relation to get an informal idea of it. As for 2, first note that $q \succ p$ implies $q \geq p$ (as stated in the linked article, too), and here's an example where the two relations differ: Consider the set $S = \{a,b,c,d\}$, so that its power set $V$ is a complete ...


1

The idea of convergence for a sequence $\{a_n\}$ is more or less the following: we say that $a_n$ converges to $a$ if for $n$ very very large, $a_n$ is very very close to $a$. How do we translate this in math? We use $\epsilon$ to say "how close" to $a$ we want to be, and $N$ to say "how far" in the sequence we need to go to stay that close. So ...


1

The statements agree when the extension is finite as you point out. However, the notion of degree is not meaningful in the context of infinite extensions. Thus, the second characterization does not exist for infinite extensions.


1

Comment promoted to answer, per request of OP. Let the vector space be ${\bf R}^2$, and let $f((a,b),(c,d))=ac$. Note that $f((0,17),(0,17))=0$.


1

The matrix can be taken to be symmetric, with $P_{ii}$ the coefficient of $x_i^2$, and $P_{ij} = P_{ji}$ half the coefficient of $x_i x_j$, in the quadratic form. Thus in your example, $P = \pmatrix{1 & 1/2 \cr 1/2 & 1\cr}$. The matrix has the same definiteness (positive definite, positive semidefinite, indefinite, negative semidefinite, negative ...


1

Let $\langle X,\le\rangle$ be a linear order. Each $x\in X$ defines a closed cofinal segment of $X$, $$[x,\to)=\{y\in X:x\le y\}$$ and an open cofinal segment $$(x,\to)=\{y\in X:x<y\}\;.$$ A subset $C$ of $X$ is cofinal if for each $x\in X$ there is a $y\in C$ such that $x\le y$. The term segment isn’t entirely standard, but it indicates that there ...



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