Why isn't the reflexive property redundant we defining an equivalence class? If $x \sim y$, then $y \sim x$ by the symmetric property. Using the transitive property we can deduce that $x \sim x$.
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An equivalence relation is defined by three properties: reflexivity, symmetry and transitivity.  Doesn't symmetry and transitivity implies reflexivity? Consider the following argument.  For any $a$ and $b$, $a R b$ implies $b R a$ by symmetry. Using transitivity, we have $a R a$.  Source: Exercise 8.46, P195 of Mathematical Proofs, 2nd (not 3rd) ed. by Chartrand et al
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Can someone give me an example for this question on reflexivity please. How is this reflexive? Can you show an example with numbers from the set?     Question 6   Let $X = \{0,1,2,3,\dotsc, 9\}$. Define a relation $\mathcal R$ from $X$ by "$x$ is related to $y$ if $x$ and $y$ gives the same remainder on division by $3$".   (a) Show that $\mathcal R$ is an equivalence relation.   Anwser:   (a) $\mathcal R$ is clearly both reflexive and symmetric, for every number has the same (unique) remainder on division by $3$ as itself and if $a$ has the same remainder on division by $3$ as does $b$, then $b$ has the same remainder on division by $3$ as $a$ also. Similarly $\mathcal R$ is transitive, for if $a\mathcal R b$ and $b\mathcal R c$, then $a$, $b$, and $c$ all have the same remainder on division by $3$, so $a\mathcal R c$ also.
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I'm following along Wikipedia page() on universal property, and this seems it should be trivial, but I couldn't finish the proof.  The definition I am working with:   The problem   So I want to prove that if $(A, u)$ and $(A', u')$ are universal morphism from $X$ to $F$, then there exists unique isomorphism $k: A \rightarrow A'$ such that $u' = F(k) \circ u$.   My attempt:   For any $B \in C$ and $f: X \rightarrow F(B)$, there exists unique $h: A \rightarrow B$ such that $f = h \circ u$.  Letting $B = A'$ and $f = u'$, we see that there exists unique $k: A \rightarrow A'$ such that $Fk \circ u = u'$.  This shows existence and uniqueness. We only need that $k$ is isomorphism, but this is the part I cannot prove.   Similarly, there exists unique $k': A' \rightarrow A$ such that $Fk' \circ u' = u$.  Plugging into each equation, we get  $$Fk \circ Fk' \circ u' = F(k \circ k') \circ u' = u'$$  $$Fk' \circ Fk \circ u = F(k' \circ k) \circ u = u$$  but this of course doesn't imply $F(k \circ k') = id$, so it seems useless.   Thanks for your help!
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I have a problem: Is it true that every equivalence between derived categories preserve their distinguished triangles? Thanks very much!
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Since $T_p G$ is isomorphic to $T_e G$ for all $p\in G$, it makes sense that each vector in $T_p G$ can be identified with a vector in $T_e G$. Hence, to make the map from $TG$ one to one, we must identify each vector in $T_e G$ with a point in $G$, but I'm having trouble saying this precisely.  Thanks so much!
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This may be a bit of a trivial question, but can one prove the reflexive, symmetric and transitive properties of equality and the transitive property of inequality of real numbers?(and if so, how? Is there a fairly straightforward, possibly algebraic method?) i.e. Is it possible to prove that $\forall\;a, b\in\mathbb{R}$ $$a=a,$$ $$\text{if } a=b \text{ then } b=a,$$  and $$\text{if } a=b \text{ and } b=c \text{ then } a=c.$$  Also, $$\text{if } a&lt;b \text{ and } b&lt;c \text{ then } a&lt;b$$ The sources that I've read so far seem to have conflicting points of view.
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For $x,y\in I:=[0,1]$ define the relation on $I$ as $x-y\in \Bbb Q$. How big (using cardinal number) is the cardinality of the equivalence class $[1/\sqrt2]$?  I have tried to solve it by finding the equivalence class but I'm not sure about the method that I can solve the question. Please, can you help me to solve it.
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Let $A$ be a ring and $S$ a multiplicative closed set. Then the localization of $A$ with respect to $S$ is defined as the set $S^{-1}A$ consisting of equivalence classes of pairs $(a, s)$ where to such pairs $(a,s), (b,t)$ are said to be equivalent if there exists some $u$ in $S$ such that  $$u(at-bs)=0$$ Now, in the  about the localization of a ring, it says that the existence of that $u\in S$ is crucial in order to guarantee the transitive property of the equivalence relation.   I've seen the proof that the equivalence relation defined above is indeed an equivalence relation, but I fail to see how crucial the existence of $u$ is. For example, why doesn't it work if we simply say that two pairs $(a,s),(b,t)$ are equivalent iff $at - bs = 0$? I tried to come up with a counterexample for such case, but failed in the attempt.
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If $X,Y$ are reflexive, symmetric, and transitive, then $X \times Y$ is an equivalence relation where ${(a,b):a\in X, b\in Y}$.    I am trying to self learn these topics. I do know what an equivalence relation is. I wanted to know how would someone prove that $X\times Y$ be an equivalence relation?
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Problem:  If a set $A$ has $n$ elements in it, how many reflexive relations can be defined on it?  My solution  Is the answer  summation of (n^2 - n)C(i)  for i=0 to n^2 -n  $$\sum_{i=0}^{n^2-n} C(n^2-n,i) = \sum_{i=0}^{n^2-n} \binom{n^2-n}i$$  How? well if i make a matrix $n\times n$  now the diagonal elements have to be selected,out of remaining $n^2-n$ any number of elements can be selected.
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I can prove it is reflexive and transitive but I cannot prove that it is antisymmetric. Help.
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Show that the logic $ \ \forall x \exists y \ P(x,y) \ $ is not logically equivalent to $ \ \exists x \forall y \ P(x,y) \ $  in the domain $ \ \{-1,0,1 \} \ $ by giving a counter example .  Answer:  Let   $ x=\{0,1 \} \\ y=\{-1,0 ,1 \} \ $  Let $ P(x,y) \ $ be the property such that $ \ x \geq y \ $   Then , $ \forall x \exists y \ P(x,y) \ $ is true.   But, $ \ \exists x \forall y \ P(x,y) \ $ is False. Because for $ x=0 \ $ , we have $ \ 0 \ngeqslant y=1 \ $.  I need confirmation of my work.
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Just need some verification for this question or another way to approach it.  Given: a~b is defined on the rationals by a~b iff $b = ak^2$ for $some$ rational number $k$.  Is it reflexive? Symmetric? Transitive?(Show the Equivalence relation)  Reflexive: Must show that a~a   $a=ak^2$  $1=k^2$ which checks out. Its Reflexive  Symmetric: Must show that a~b and b~a  a~b: $b=ak^2$, $\frac ba$=$k^2$  b~a: $a=bk^2$, $\frac ab$=$k^2$  Not symmetric.  Transitive:   a~b and b~c than we must show that a~c  a~b: $b=ak^2$,  b~c: $c=bk^2$  a~c: $c=ak^2$  This is were I am sorta stuck not sure if I am allowed to do the following:  since $c=ak^2$ and $c=bk^2$ I can find that a=b?not sure any help would be appreciated.  Noticing that if zero is excluded then we would have an equivalence relation. However, when zero is included would symmetric fail and both reflexive and transitive still hold?
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I came across this long proof on this site:    But I would like to know whether my direction can work.  Say we want to find the cardinality of all equivalence relations in $\mathbb{N}$. Since it is a subset of all relations in $\mathbb{N}$, I conclude it has a cardinality smaller or equal to $\aleph$. Now, define an injective function from $P(\mathbb{N})$ to the set of equivalence relations by matching each subset of $\mathbb{N}$ with the identity relation (which is an equivalence relation in $\mathbb{N}$.  Therefore the cardinality of all equivalence relations in $\mathbb{N}$ is greater or equal to $\aleph$ and using CSB we get the desired result.  Seems legit?
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I have a doubt. I need help.  Can the basic axioms of equality be presented as "open axioms"?  I) (reflexivity) $\qquad x = x$              II)  (Substitutivity) $\qquad (x = y) \to \big(F (x, x) \to F (x, y)\big)$
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Let $X$ and $Y$ be homeomorphic topological spaces, connected by the homeomorphism $f : X \rightarrow Y$. Let $\sim$ be an equivalence relation on $X$ and $\approx$ be an equivalence relation on $Y$.  I would think that there would be some structure that I could place on the equivalence relations $\sim$ and $\approx$ that would allow me to construct, using the homeomorphism $f$, a homeomorphism $\bar{f}$ from $X/\sim$ to $Y/\approx$. I suspect that that relationship is $a \sim b \leftrightarrow f(a) \approx f(b), \forall a, b \in X$.  That is, I suspect that "equivalent" quotients of homeomorphic spaces are homeomorphic, but I don't know exactly how to formulate this or prove it.  (This result, by the way, to me, seems to hinge on the existence of a "reverse" universal quotient property: juts as functions $f : X/\sim \rightarrow Y$ correspond uniquely to a subset of functions $f : X \rightarrow Y$, I suspect that functions $f : X \rightarrow Y/\approx$ correspond uniquely to some subset of functions $f : X \rightarrow Y$, but I don't know how this works either.)
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Let $S$ be the relation on $\mathbb{N} \times \mathbb{N}$ defined by $(a,b)S(c,d)$ if and only if $ad=bc$. Prove this is an equivalence relation on $\mathbb{N} \times \mathbb{N}$.  I think I've found this to be reflexive and symmetric, but I'm stuck on transitivity. Can someone check my work so far and assist with testing transitivity?  Reflexive: Let $(x,y)S(x,y)$. Then $xy = yx$. So $S$ is reflexive. Symmetric: Suppose $(a,b)S(c,d)$. Then, $ad = bc$. Therefore, $da = cb$ and $cb = da$. Therefore, $(c,d)S(a,b)$. Thus, $S$ is symmetric.  Transitive: Suppose $(a,b)S(b,c)$. Then $ac = bb$.   This is where I'm stuck. Any ideas?