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35

Consider the tropical semiring $(\mathbb R\cup\{\infty\},\oplus,\otimes)$, where $x\oplus y=\min\{x,y\}$ and $x\otimes y=x+y$. The "additive" identity, meaning the identity element for the $\oplus$ operation, is $\infty$. That's pretty far away from $0$! The number $0$ is present, but it is not the additive identity in this structure. For example, $0\oplus ... 21 In the integers under the operation$*$defined by$n * m = n + m - 1$,$1$acts as the additive identity. Of course, this is identical to the ordinary additive group structure on the integers except for relabeling. But there are occasional situations where people find themselves using this operation, usually when there are historical off-by-one errors to ... 10 A lightswitch has two values, off and on. Assemble two lightswitches in parallel. You now have the following addition table: $$\begin{array}{c|cc} + & \textrm{off} & \textrm{on} \\ \hline\\ \textrm{off} & \textrm{off} & \textrm{on} \\ \textrm{on} & \textrm{on} & \textrm{on} \end{array}$$ Obviously,$\textrm{off} + \textrm{on} = ...

6

I am surprised no one mentioned elliptic curves with its "usual" elliptic-curve-addition. You pick two points $a$ and $b$ on an elliptic curve. Then $a+b$ is defined as the negative of the third point of intersection between the elliptic curve and the straight line connecting $a$ and $b$. In this case, the additive identity is denoted $O$ which is the point ...

4

\begin{align*} & & x &\in (A \cup B) \cap C \\ &\iff & x &\in (A \cup B) \wedge x \in C \\ &\iff & (x &\in A \vee x \in B) \wedge x \in C \\ &\iff & (x &\in A \wedge x \in C) \vee (x \in B \wedge x \in C) \\ &\iff & (x &\in A \cap C) \vee (x \in B \cap C) \\ &\iff ... 4\varnothing\ne\{\varnothing\}$. The second is the only element of$\{\{\varnothing\}\}$. Do you see$\varnothing$by itself listed in there? No. When we say every set contains the empty set, we mean as a subset, not as a member or element, so we are saying that$\varnothing\subseteq X$not$\varnothing\in X$. Do you think$\varnothing\in\varnothing$for ... 4 There are two basic ideas: Given$X$, if it is finite then pick any bijection with$\Bbb Z/(n)$, and you have a finite group; otherwise consider$\Bbb Z[X]$, the ring of polynomials whose free variables are elements of the set$X$. We can prove, using the axiom of choice, that$\Bbb Z[X]$has the same cardinality as$X$. Therefore there exists a bijection ... 4 A module is an abelian group. (It's more useful to think of a module as the analog of a vector space, but with the set of scalars coming from a ring instead of a field. Usually, one arrives at this notion of a module in terms of "the action of a ring on a set" where the set is a module.) A monoid is a relaxation of the definition of a group. A monoid has ... 3 There is a chain of forgetful functors which progressively forget the various operations in the structure: $$\mathrm{Mod_R}\to\mathrm{Ab}\to\mathrm{AbMon}\to\mathrm{Set}$$ The interesting thing is that you can go in the opposite direction too with free functors $$\mathrm{Set}\to\mathrm{AbMon}\to\mathrm{Ab}\to\mathrm{Mod_R}$$ Each forgetful functor$U$is ... 3 The word contains is tricky, as it can mean both has as an element and is a superset of, depending on the context. Let's avoid it here. Every set$A$is a superset of$\emptyset$. But this is a theorem, not an axiom: there is a reason why this is true. The definition of the subset relation is:$B$is a subset of$A$(written$B\subseteq A$) ... 3 Long time since I did math but the easisest example i can I think if is modular arithmetic. For a≡b (mod n), n is an additive identity. For instance the modular arithmetic for determining which day of the week has 2 additive identities 0 and 7. Informally, Monday + 0 days = Monday, Monday + 7 days = Monday. 3 Hint: Set$\{\varnothing\}$, sometimes written as$\{\{\}\}$is "an empty box in a box". In this wording$\{\varnothing,\{\varnothing\}\}$is a box that contains two boxes, one empty, while the second contains yet another empty box. Yes, you are correct. We have$\varnothing \notin \{a,b,c\}$, still$\varnothing \subset \{a,b,c\}$. Also,$\varnothing \in ...

3

You will probably have to leave number systems if you want to get away from 0 being the additive identity. For instance, if you take the + operation to be permutations, then you would have a non-number additive identity, which would be the identity permutation (i.e. don't permute anything). You could also take the operation of symmetries in the plane, ...

2

Each subset $S_x \subseteq \{0,1,\dots, 9\}$ can be mapped to a vector $\vec{x} \in \{0,1\}^{10}$ by defining $x_i = 0$ if $i \notin S_x$ and $x_i = 1$ otherwise, for $i = 0, \dots, 9$. In terms of such vectors, the condition $|S_x \Delta S_y| \leq 2$ then translates to a Hamming distance of at most $2$ between the vectors $\vec{x}$ and $\vec{y}$. So if ...

2

Consider the sets $U = [ 0 , \omega+1 ) \subseteq \omega_1$ and $V = [ 0 , \frac{1}{2} )$. Clearly $U$ and $V$ are open in $\omega_1$, $[0,1)$, respectively, and so $U \times V$ is open in $\omega_1 \times [0,1)$ with the product topology. However this set is not open in the dictionary order topology. Note that $\langle \omega , 0 \rangle \in U \times V$. ...

2

When we write $(a,b)$ is a neighbourhood for $c\in\mathbb R$, it is indeed implied that $c$ is fixed. On the other hand, it is correct to say that for any $c\in(a,b)$, the interval $(a,b)$ is an open neighbourhood for $c$. Finally, neighbourhoods are usually defined to be open sets. But sometimes it is necessary to be more general, and therefore another ...

2

I think that your confusion can be traced to a simple grammatical error: The definition should read: A neighborhood of some element $c$ is an open interval $(a,b)$ such that $a<c<b$. Neighborhoods aren't unique - typically each point has infinitely many of them! So we don't talk about the neighborhood of a point, we just say that some ...

2

Yes, very much. Because when switching to $|A_i|$ you need to effectively choose canonical representatives from each equivalence class, and bijections from $A_i$ to that set. Neither of these processes is well-defined without the axiom of choice. We can have the following situation: There exists a set $S$ which can be partitioned into countably many ...

2

Let's just take the case where $V_0 = \varnothing$. When we do this it follows easily that $V_{n+1} = \mathcal{P} ( V_n )$ for all $n < \omega$. So let's look at the first few levels of this hierarchy: $V_0 = \varnothing$; $V_1 = \mathcal{P} ( V_0 ) = \mathcal{P} ( \varnothing ) = \{ \varnothing \}$; $V_2 = \mathcal{P} ( V_1 ) = \mathcal{P} ( \{ ... 2 Ok, I see that$B$is just the set of singletons from$\mathbb{Z}$. It seems that$C$is the collection$\left\{\{x,x+1\}|x\in\mathbb{R}\right\}$, so you're right about those two cardinalities. The third one,$D$, should include any subset of$\mathbb{R}$that has diameter less than$1$. (It also includes sets such as$(0,1)$with diameter$1$, but those ... 2 Intuitively, it's because while functions have to be well-defined, they're allowed to be non-injective, so if you're placing arrows between the elements of two objects, you have a lot more freedom in choosing where to make your arrows point than where you can choose where to make them point from. To put it more succinctly, you can find more ways to send ... 2 This is a question of topology, in which these are taken to be axioms defining open and closed sets. To see why they are reasonable, take a few examples:$A_n = (-\frac{1}{n},\frac{1}{n})$is an open set for every$n$. What happens if we take an infinite intersection? $$\bigcap_{n=1}^\infty A_n = \{0\}$$. If we allow infinite intersections of open sets to ... 2 1.a. is not reflexive because it does not include$(1,1)$. It is symmetric. It is not transitive because, although it includes$(1,3)$and$(3,1)$, it does not include$(1,1)$1.b. Is reflexive and is not symmetric, for precisely the reasons you said. It is also transitive because there is no counterexample to it being transitive. There are only two that ... 2 The Cartesian product of two sets,$P, Q$is defined as the set $$P\times Q = \{(p, q)\mid p\in P, \;\;q \in Q\}$$ How many$p$'s are in$P$? Given:$\;m$How many$q$'s are in$Q$. Given:$\;n$For each$p \in P$there are$n$elements in$Q$, so, for example $$(p_1, q_1), (p_1, q_2), \ldots, (p_1, q_n) \in P\times Q.$$ We get a list of$\bf n$ordered ... 1 As you said that ordering is important, so 1) is obviously not the correct solution because any instruction will belong to b and ordering is not important in that case. 2) is absolutely fine but what you have written that,$0\le x_{i}\le k$, I think you have done a mistake here instead of$x_{i}$you only have to write$i. 1 In the first problem, note that when you get to: \begin{align} ...&= A \cap (C^\complement \cup D^\complement) \cup B \cap (C^\complement \cup D^\complement)\\ \\ & = (A \cap C^\complement) \cup (A \cap D^\complement) \cup (B \cap C^\complement) \cup (B \cap D^\complement)\end{align}\tag{1} (You mention De Morgan's when moving from ... 1 As you said, you think of\mathbb R$as a line and$\mathbb R\times\mathbb R$as a plane. Well, we think of$\mathbb R \sqcup \mathbb R$as two disjoint lines. Of course already thinking of$\mathbb R$as a line implies a topology on$\mathbb R$at least, as well as thinking of$\mathbb R\times\mathbb R$as a plane implies a topology, after all$\mathbb ...

1

I think you meant "..., with $(x,y)\in R$ and $(y',z)\in R$." but with that correction your reasoning seems fine to me. We have both that $R$ is vacuously transitive (because there are no elements $\alpha$ such that $(\alpha,\cdot)$ and $(\cdot,\alpha)$ are in $R$) and that $S$ is not transitive for the reason you gave.

1

$g$ is surjective because for any $y\in \Bbb N$ we have $g(2y)=y$, hence the image of $g$ covers the whole $\Bbb N$. The domain of a function is supposed to be the whole space: if we're talking about a function $h:X\to Y$, then by definition $h(x)$ is defined for all $x\in X$. Thus all functions $f_m$ are defined for all $k$. Finally, $1+\sum_{m\le ... 1 Here's a fun exercise. Start with$V_0=\varnothing$. Now prove that for every$n$,$V_n$has only finitely many elements. Conclude, if so, that every set in$V_n$is finite as well. Therefore every set in$V_\omega$is finite, but$V_\omega\$ itself is infinite. In order to produce an infinite set as an element of the hierarchy, we have to go one more step, ...

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