求高手翻译！
在沪江关注考试英语的沪友瑟雅姐遇到了一个关于的疑惑，已有1人提出了自己的看法。
知识点疑惑描述：
His father is still alive and well.
最佳知识点讲解
知识点相关讲解
他的父亲依然活得很好。
—— Shu9304
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self providedhot......回答时间：[]Copyright & 2017BINARY RELATIONS; EQUIVALENCES　　　　In our discussion of ordered sets we have already encountered the ides of a (binary) relation on a set X, by which we simply mean a subset ρ of the Cartesian product X × X. At this stage it is convenient to develop the theory of relations in a somewhat more general and abstract way. Intuitively we think of elements x and y for which (x, y) ∈ ρ as being related, and we frequently prefer to write x ρ y instead of (x, y) ∈ ρ. The empty subset &O of X × X is included among the binary relations on X; other special relations worthy of mention are the universal relation X × X, in which everything is related to everything else, and the equality relation　　　　 1X = {(х,х) : х∈X}, (1.4.1)　　　　also known as the diagonal relation, in which two elements are related if and only if the are equal.　　　　Let us denote the set of all binary relations on X by βх. A binary operation o is defined onβх by the rule that, for allρ,σin βх,　　　　ρoσ={ (x, y) ∈ X × X :( z ∈X)(x, z) ∈ ρand(z, y) ∈σ}. (1.4.2)　　　　It is easy to see that, for all ρ,σ,τinβх,　　　　ρ σ ρoτ σoτand τoρ τoσ. (1.4.3)　　　　It is easy to see also that the operat for allρ,σ,τinβх,　　　　 (ρoσ) oτ=ρo(σoτ),　　　　for　　　　 (x, y)∈(ρoσ) oτ　　　　 ( z∈X)(x, z)∈ρoσand (z, y)∈τ,　　　　 ( z∈X)( u∈X)(x, u) ∈ ρ, (u, z) ∈σand (z, y)∈τ,　　　　 ( u∈X)(x, u) ∈ ρand (u, y) ∈σoτ,　　　　 (x, y) ∈ρo(σoτ).　　　　We have proved　　　　Proposition 1.4.1 Let βх be the set of all binary relations on a set X, and define the operation o on βхby (1.4.2). Then (βх, o) is a semigroup.　　　　Whilst we shall not normally revert to simple multiplicative notation when discussing the semigroup (βх, o) , we shall allow ourselves to writeρ²,ρ³, etc, instead of ρoρ,ρoρoρ, etc.　　　　For each ρ∈βх we define the domain dom ρby　　　　 dom ρ={x∈X: ( y ∈X)(x, y) ∈ ρ}. (1.4.4)　　　　And the image im ρ by　　　　 im ρ={y∈X: ( x ∈X)(x, y) ∈ ρ}. (1.4.5)　　　　It is immediate that, for allρ,σinβх,　　　　 ρ σ domρ dom σand im ρ imσ. (1.4.6)　　　　For each x in X andρinβхwe define a subset xρof X by 　　　　 xρ={y ∈ X:(x, y) ∈ρ} ; (1.4.7)　　　　thus xρ≠&O if and only if x∈dom ρ. If A is a subset of X we define　　　　 Aρ=∪{aρ:a∈ A}. (1.4.8)　　　　For each ρinβх, we define ρ , the converse of ρ, by　　　　 ρ ={(x, y) ∈X× X:(y, x) ∈ρ}. (1.4.9)　　　　Certainlyρ ∈βх,and it is easy to see that, for all ρ,σ,ρ1,ρ2,…ρn inβх,　　　　(ρ ) = ρ, (1.4.10)　　　　 (ρ1ορ2ο…ορn) = ρn ο…ορ2 ορ1 , (1.4.11)　　　　 ρ σ ρ σ . (1.4.12)　　　　Notice also that　　　　 dom (ρ ) = im ρ, im(ρ ) = dom ρ, (1.4.13)　　　　and that　　　　 xρ ≠&O if and only if x∈imρ.　　　　An element & ofβхis called a partial map of X if |x&|=1 for all x in dom & , that is, if, for all x , y1, y2 in X,　　　　 [(x, y1)∈& and (x, y2)∈&] y1=y2. (1.4.14)　　　　It will not conflict at all with (1.4.7) if we decide in such a case to let x& denote the unique element y such that (x, y)∈& (rather than the set consisting of that element). Notice that the condition (1.4.14) is fulfilled (vacuously) by the empty relation &, which is therefore included among the partial maps.　　　　If &, ψare partial maps of X such that & ψ. We sometimes say that & is a restriction of ψ, or thatψis an extension of & .If in these circumstances, dom & =A domψ, then we denote & by ψ|A (ψrestricted to A).　　　　Proposition 1.4.2 The subset Px of βхconsisting of all partial maps of X is a subsemigroup of βх.　　　　Proof Let &, ψ∈Px , and suppose that (x, y1), (x, y2)∈&οψ.Then there exist z1, z2 in X such that　　　　 (x, z1)∈&, (z1, y1)∈ψ, (x, z2)∈&, (z2, y2)∈ψ. 　　　　The condition (1.4.14) on & implies that z1 = z2, and then the same condition onψimplies that y1 = y2 . Thus &οψ∈Px .　　　　It is important to note that the converse & of a partial map & need not be a partial map. For example, if X={1,2}, then & ={(1,1), (2,1)} is a partial map, but & is not.　　　　In view of proposition 1.4.2 we can talk of (Px, o) as the semigroup of all partial maps of X. The composition law o in this semigroup is in fact a fairly natural composition law for partial maps.　　　　Proposition 1.4.3 If &, ψ∈Px , then　　　　dom(&οψ)=[im&∩domψ] & ,　　　　im(&οψ)= [im&∩domψ]ψ,　　　　And　　　　 ( x∈dom(&οψ)) x(&οψ)=(x&)ψ.　　　　Proof Before proving this, we illustrate it in a diagram as follows:　　　　　　Suppose first that x∈dom(&οψ) . Then there exist y and z in X such that (x, z)∈& , . Clearly z∈im&∩dom , and (z, x)∈& .　　　　Hence　　　　 x∈z& [im&∩domψ] & .　　　　Conversely, suppose that x∈[im&∩domψ] & . Then there exists z in im&∩domψsuch that x∈z& , that is, such that (x, z)∈& . Since z∈domψ, there exists y in X such that (z, y)∈ψ.Hence (x, y)∈&οψand so x dom (&οψ). Thus　　　　 dom(&οψ)=[im&∩domψ] & ,　　　　as required. The proof that im(&οψ)= [im&∩domψ]ψis similar.(Notice that as yet we have used no special properties of partial maps. The characterizations established for dom (&οψ) and im (&οψ) apply equally well to arbitrary relations.)　　　　To complete the proof, notice that (x, y)∈&οψif and only if there exists z in X such that (x, z)∈& and (z, y)∈ψ. Since &, ψand &οψare partial maps, we have z=x&, y=zψ, and y=x(&οψ). Hence　　　　 x(&οψ)=y= zψ=(x&)ψ,　　　　exactly as required.　　　　A partial map & is called a map, or a function, if dom & =X. Thus a relation & on X is a map if and only if |x&|=1 for every x in X. If & andψare maps, it is easy to see that &οψis again a map, and that the operation (1.4.2) coincides with ordinary composition of maps. To put it formally, we have.　　　　Proposition 1.4.4 The setΤx of all maps from X into itself is a subsemigroup of (βх, o).　　　　Once again it is important to note that the converse & of a map need not be a map. Indeed we have the following easily proved result:　　　　Proposition 1.4.5 Let X be a non-empty set.　　　　(1) If &∈Px then & ∈Px if and only if & is one-one.　　　　(2) If &∈Τx then & ∈Τx if and only if & is bijective (that is to say, & is both one-one and onto).　　　　In Section 1.3 we encountered relations, namely order relations, that are reflexive, anti-symmetric and transitive. We can now express these properties very compactly as follows. A relation ρon a set X is 　　　　 reflexive is and only if 1x ρ,　　　　 anti-symmetric if and only if ρ∩ρ =1x ,and　　　　 transitive if and only ifρoρ ρ.　　　　We now define an equivalenceρon a set X to be a relation that is reflexive, transitive and symmetric, by which we mean that　　　　 ( x, y∈X)(x, y)∈ρ (y, x)∈ρ.　　　　In compact from this property is expressed asρ ρ . Notice that, by (1.4.12) and (1.4.10) it then follows that ρ ρ; thus the symmetry condition can equally well be expressed asρ =ρ. On the same theme, if ρis an equivalence, then by (1.4.3) we can deduce that　　　　 ρ－1xoρ ρoρ;　　　　thus the transitivity condition can be replaced byρoρ=ρ.　　　　If ρis an equivalence on X then, by (1.4.6),　　　　 domρ dom 1x =X, imρ im 1x =X;　　　　hence dom ρ= imρ= X..　　　　A familyπ={Ai: i∈I}of subsets of a set X is said to from a partition of X if　　　　（P1）each Ai is non-　　　　(P2) for all i, j in I, either Ai = A or Ai ∩ Aj = &O;　　　　(P3) ∪{Ai: i∈I} = X.　　　　On the face of it, the notions of ‘partition’ and ‘equivalence’ are quite different, but it fact they are closely related. The proof of the following proposition is routine and is omitted.　　　　Proposition 1.4.6 Letρbe an equivalence on a set X. Then the family 　　　　 Φ(ρ)={xρ:x∈X}　　　　of subsets of X is a partition of X.　　　　Conversely, if π= {Ai: i∈I}is a partition of X, then the relation　　　　 Ψ(π)={(x, y)∈X × X: ( i∈I)x, y∈Ai}　　　　is an equivalence on X.　　　　For every equivalence ρon X, Ψ(Φ(ρ))=ρ, and for every partitionπof X, Φ(Ψ(π)) =π.　　　　If ρis an equivalence on X, we shall sometimes write x ρ y or x≡y(modρ) as alternatives to (x, y)∈ρ. The sets xρthat from the partition associated with the equivalence are called ρ-classes, or equivalence classes. The set of ρ-classes, whose elements are the subsets xρ, is called the quotient set of X by ρ, and is denoted by X∕ρ. In the next section we shall have occasion to examine the natural map ρ (read ‘ρnatural’) from X onto X∕ρdefined by　　　　 xρ =xρ (x∈X). (1.4.15)　　　　An important connection between maps and equivalences is given by　　　　Proposition 1.4.7 If &: X →Y is a map, then &ο& is an equivalence.　　　　Proof The easiest way to see this is to note that　　　　&ο& ={(x, y)∈X × X: ( z∈X) (x, z)∈&, (y, z)∈&}　　　　 ={(x, y)∈X × X: x&=y&}.　　　　It is then clear that &ο& is reflexive, symmetric and transitive.　　　　We call the equivalence &ο& the kernel of & and write &ο& = ker & .Notice that ker ρ =ρ.　　　　If {ρi: i∈I} is a non-empty family of equivalences on a set X, then, as may be verified in a routine manner, ∩{ρi: i∈I} is again an equivalence.　　　　If R is any relation at all on X—even the empty relation will do—then the family of equivalences containing R is non—empty, since X × X hence the intersection of all the equivalences containing R is an equivalence, the unique minimum equivalence on X containing R. We call it the equivalence generated by R, and denote it by R .　　　　It is frequently necessary to be able to describe R for a given R, and the foregoing general description is not particularly useful. It is necessary therefore to develop an alternative description. First, let S be a relation on X such that 1x S—a reflexive relation, in fact. Then we have　　　　 S SοS SοSοS … ,　　　　which we can write in simpler notation as 　　　　 S S² S³ …　　　　The relation　　　　 S =∪{S : n≥1} (1.4.16)　　　　is called the transitive closure of the relation S, a term that is justified by the following lemma:　　　　Lemma 1.4.8 For every reflexive relation S on a set X, the relation S defined by (1.4.16) is the smallest transitive relation on X containing S.　　　　Proof First, S is transitive. Suppose that (x, y), (y, z)∈S . Then there exist positive integers m and n such that (x, y)∈S and (y, z)∈S . It follows that　　　　 (x, z)∈S οS =S S .　　　　It is clear that S contains S = S.　　　　Finally, is T is a transitive relation containing S, then　　　　 S²=SοS TοT T,　　　　and more generally S T for all n≥1. Hence S T.　　　　We now have　　　　Proposition 1.4.9 For every relation R on a set X,　　　　 R =[R∪R ∪1x] .　　　　Proof From lemma 1.4.8 we see that the relation E=[R∪R ∪1x] is transitive and contains R. Since　　　　 1x R∪R ∪1x E.　　　　E is also reflexive. Certainly the relation S =R∪R ∪1x is symmetric, and it follows that, for every n in N,　　　　 S =(S ) =(S ) ,　　　　the second equality being a specialization of (1.4.11). Hence S is symmetric. It now follows that E = S is symmetric, since　　　　 (x, y)∈E ( n∈N)(x, y)∈S 　　　　 ( n∈N)(y, x)∈S 　　　　 (y, x)∈E.　　　　We have shown that E is an equivalence relation containing R.　　　　Suppose now thatσis an equivalence relation containing R. Then 1x σ, and R σ .Hence　　　　 S=R∪R ∪1x σ.　　　　Moreover,　　　　 SοS σοσ=σ,　　　　and more generally S σfor all n≥1. It follows that E=S σ. We have shown that E=[R∪R ∪1x] is the smallest equivalence on X containing R. Thus　　　　R =[R∪R ∪1x] 　　　　as required.　　　　In more down-to-earth terms, Proposition 1.4.9 can be rewritten thus:　　　　Proposition 1.4.10 If R is a relation on a set X and R is the smallest equivalence on X containing R, then (x, y)∈R if and only if either x=y or, for some n in N, there is a sequence of transitions　　　　 x=z1→z2→…→zn=y　　　　in which, for each i in {1,2,….n-1}, either (zi, zi+1)∈R or (zi+1,zi)∈R .　　　　1.5 CONGRUENCES　　　　Let S be a semigroup. A relation R on the set S is called left compatible (with the operation on S) if　　　　 ( s, t, a∈S) (s, t)∈R (as, at)∈R,　　　　and right compatible if　　　　 ( s, t, a∈S) (s, t)∈R (sa, ta )∈R　　　　It is called compatible if　　　　 ( s, t, s’, t’∈S) [(s, t)∈R and (s’, t’)∈R] (ss’, tt’)∈R.　　　　A left[right] compatible equivalence is called a left[right] congruence. A compatible equivalence relation is called a congruence.　　　　Proposition 1.5.1 A relation ρon a semigroup S is a congruence if and only if it is both a left and a right congruence.　　　　Proof Suppose first that ρis a congruence. If (s, t) ∈ ρ and a∈ S then (a, a) ∈ ρ by reflexivity and so (as, at) ∈ ρand (sa, ta) ∈ ρ by compatibility. Thus ρ is both left and right compatible.　　　　Conversely, suppose that ρis both a left and a right congruence, and let (s, t), (s’, t’) ∈ ρ. Then (ss’, ts’) ∈ ρby right compatibility and (ts’ tt’) ∈ ρby left compatibility. Hence (ss’, tt’) ∈ ρby transitivity. Thusρis a congruence.　　　　Exercise 11 below explores in more detail the relationship between left and right compatibility on the one hand and compatibility on the other.　　　　If ρis a congruence on a semigroup S then we can define a binary operation on the quotient set S∕ρin a natural way as follows:　　　　 (aρ)(bρ)=(ab)ρ. (1.5.1)　　　　This is well-defined precisely because ρis compatible: for all a, a’, b, b’ in S,　　　　aρ=a’ρand bρ (a, a’)∈ρ and (b, b’)∈ρ　　　　 (ab, a’b’)∈ρ　　　　 (ab)ρ=(a’b’)ρ.　　　　The operation is easily seen to be associative, and so S∕ρis a semigroup.　　　　We have proved part of the following theorem:　　　　Theorem 1.5.2 Let S be a semigroup, and let ρbe a congruence on S. Then S∕ρis a semigroup with respect to the operation defined by (1.5.1), and the map ρ from S onto S∕ρgiven by (1.4.15) is a morphism.　　　　Now let T be a semigroup and let &: S→T be a morphism. Then the relation　　　　 ker &=&ο& ={(a, b)∈S × S: a&=b&}　　　　Is a congruence on S, and there is a monomorphism α: S∕ker & →T such that im α= im & and the diagram commutes.　　　　
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