Let's define equivalence relation on group $\mathbb{R}$ by setting $x\sim y$ iff $x-y\in\mathbb{Z}$ by all $x,y\in\mathbb{R}$.  What is the quotient topology?  For example if we set $x=\sqrt{2}$ and $y=\sqrt{2} + 2$, then $x-y\in\mathbb{Z}$.  So this looks similar to the case where you replace $\mathbb{Z}$ with $\mathbb{Q}$, but $\mathbb{Z}$ is not dense in $\mathbb{R}$. In that case the quotient topology is indiscrete.  Could someone please open up this a bit for me?
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There is this example at the Wikipedia article on Quotient spaces (QS):     Consider the set $X = \mathbb{R}$ of all real numbers with the ordinary topology, and write $x \sim y$ if and only if $x−y$ is an integer. Then the quotient space $X/\sim$ is homeomorphic to the unit circle $S^1$ via the homeomorphism which sends the equivalence class of $x$ to $\exp(2πix)$.   I understand relations, equivalence relation and equivalence class but quotient space is still too abstract for me. This seems like a simple enough example to begin with.  I understand (sort of) the definition but I can't visualize. And by this example and others, there is a lot of visualizing going on here! torus, circles etc.
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I am working on this exercise for an introductory Real Analysis course:     Show that |$\mathbb{R}$| = |$\mathbb{R}^2$|.   I know that $\mathbb{R}$ is uncountable. I also know that two sets $A$ and $B$ have the same cardinality if there is a bijection from $A$ onto $B$. So if I show that there exists a bijection from $\mathbb{R}$ onto $\mathbb{R}^2$ then I beleive that shows that |$\mathbb{R}$| = |$\mathbb{R}^2$|.   Let $x_i \in \mathbb{R}$, where each $x_i$ is expressed as an infinite decimal, written as $x_i = x_{i0}.x_{i1}x_{i2}x_{i3}...,$. Each $x_{i0}$ is an integer, and $x_{ik} \in \left \{ 0,1,2, 3, 4, 5, 6, 7, 8, 9 \right \}$. Then, let  $$f(x_i)=(x_{i0}.x_{i1}x_{i3}x_{i5}... ,x_{i0}.x_{i2}x_{i4}x_{i6}...)$$  What should I do to show that $f: \mathbb{R} \to \mathbb{R}^2$ is an injective function? Any suggestions or help with the question would be appreciated.
120588
In general, I struggle working with quotients of polynomial rings, so I was hoping someone might be able to help with the following exercise:  Show that $\mathbb{Q}[x]/(x^2-2x)$ is isomorphic to $\mathbb{Q}[x]/(x^2-1)$, but the latter is not isomorphic to $\mathbb{Q}[x]/(x^2)$.  Mostly I want to understand what these quotients all &quot;look like.&quot; We did few examples in class and few homework problems with such quotients, and the ones we do work with are much simpler (typically involving $\mathbb{Z}_2$ or $\mathbb{Z}_3$ or $\mathbb{R}$ instead of $\mathbb{Q}$). Any help is appreciated, thank you!
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I still struggle with elementary knowledge about quotient rings.  For example, why can I just write down the second following isomorphism:  $$\frac{\mathbb{C}[x, y, z]}{(x^2 + y^2 - 1, y \pm iz)} \simeq \frac{\frac{\mathbb{C}[x, y, z]}{(y \pm iz)}}{\frac{(x^2 + y^2 - 1,y \pm iz)}{(y \pm iz)}} \simeq \frac{\mathbb{C}[x, y]}{(x^2 + y^2 - 1)} $$  I know the first isomorphism is right by the third isomorphism theorem. But why can I write down the second one?
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It is very elementary to show that $\mathbb{R}$ isn't homeomorphic to $\mathbb{R}^m$ for $m&gt;1$: subtract a point and use the fact that connectedness is a homeomorphism invariant.  Along similar lines, you can show that $\mathbb{R^2}$ isn't homeomorphic to $\mathbb{R}^m$ for $m&gt;2$ by subtracting a point and checking if the resulting space is simply connected. Still straightforward, but a good deal less elementary.  However, the general result that $\mathbb{R^n}$ isn't homeomorphic to $\mathbb{R^m}$ for $n\neq m$, though intuitively obvious, is usually proved using sophisticated results from algebraic topology, such as invariance of domain or extensions of the Jordan curve theorem.  Is there a more elementary proof of this fact? If not, is there intuition for why a proof is so difficult?
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Here's the problem: if $S = \mathbb{Q}[x]/(x^2)$, compute the group  of automorphisms $\mbox{Aut}(S)$.   First off, I have the solution and I know it's isomorphic to the multiplicative group $\mathbb{Q}\backslash\{0\}$, so I'm trying to understand it. It uses what is called the substitution principle and the fact that the restriction of any endomorphism $\phi: \mathbb{Q}[x]\to \mathbb{Q}[x]$ to $\mathbb{Q}$ is the identity. It says that these two facts implies that the endomorphisms of $\mathbb{Q}[x]$ are in 1-1 correspondence with $h \in \mathbb{Q}[x]$, where the endomorphism corresponding to $h$ is $\phi_h : \mathbb{Q}[x] \to \mathbb{Q}[x]$ given by $\phi_h(f) = f(h)$ for $f \in \mathbb{Q}[x]$. I guess I just don't understand how it applied the substitution principle? It's probably obvious, but for some reason I am not getting it.  An explanation would be greatly appreciated! Or if you have a different solution, that'd be cool too. Thanks!!
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$(\Bbb R,+)$ is a topological group. Is there any other group structure on $\Bbb R$ such that it is still a topological group and this group is not isomorphic to $(\Bbb R,+)$ ?  Refer to  for the general problem. In that problem I have conjectured that there is no such group however I couldn't prove it.
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In my textbook it is said that the quotient space $D^n/S^{n-1}$ is homeomorphic to $S^n$. I can imagine it for $n=2$, but fail to make a mathematical proof for any dimension.  Can anyone provide a rigorous proof?
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Let $X = \mathbb{R}$ and $Y = \{x \in \mathbb{R} :x ≥ 1\}$, and define $G : X → Y$ by $$G(x) = e^{x^2}.$$ Prove that $G$ is onto.  Is this going along the right path and if so how do get the function to equal $y$?  $G: \mathbb{R} \to\mathbb{N}_1$. Let $y$$\in $$\mathbb{N_1}$. claim: $\sqrt{\ln y}$ maps to $y$. Does $\sqrt{\ln y}$ belong to $\mathbb{N_1}$? Yes because $y \in \mathbb{N_1}$, $G( \sqrt{\ln y})=e^{(\sqrt{\ln y})^2}$.
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Assuming the following: let $X$ be a set with two elements, $X$ = {$a,b$}.  what are all the possible topologies on $X$ ?  The answer I've come up with is: $$\tau_{T}=\{ \emptyset, X\} $$ $$ \tau_{a}=\{\emptyset,\{a\},X\} $$ $$\tau_{b}=\{\emptyset,\{b\},X \} $$ $$ \tau_{D}=\{\emptyset,\{a\},\{b\},X\}$$  My question is, are these correct? and are there any more?  thanks in advance
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prove that $a \equiv b \mod m$ is an equivalence relation on the integers  I believe there are 3 properties that it must meet to prove and equivalence relationship. Any reference material would be greatly appreciated
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Define $x\sim y$ means 5 divides $(x - y)$ for $x$ and $y$ integers. Show that is an equivalence relation.  Equivalence relation means it satisfies reflexity, symmetry, and transitivity.  reflexive: $x\sim x$ means 5 divides $x$  symmetry: $x \sim y \rightarrow y\sim x$ means 5 divides $x - y$ and 5 divides $y - x$: $$5/(x - y) = 5/(y - x)$$ so symmetry is satisfied.  I am not sure if I am right here and I am lost on how to prove it is transitive any suggestions would be greatly appreciated
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I started learning topology recently, so I'm very new to this.    The question is as follows:     Given the set $A = \{ (x,y)\;;\;x, y\in\mathbb{Q} \}=\mathbb{Q}^2$, show that the complement set $A^c=\mathbb{R}^2\setminus A$ is polygonally connected.   As an attempt on this question, I plan to show that the set $A^c$ is connected, thus using a theorem, I can say $A^c$ is polygonally connected.  But here is the part that I got confused: By definition, the set $A^c$ is connected iff it cannot be written as the union of $2$ nonempty separated sets.    In this case, I believe $A^c$ is the set of all $(x,y)$ where $x,y$ belong to $\mathbb R-  \mathbb Q$.  But how can I express the fact that $A^c$ is not the union of $2$ nonempty separated sets ?   Based on my knowledge, I know both sets $\mathbb Q$  and $\mathbb R-\mathbb Q$ are dense, but how is this helpful to solving this problem?  I think a reason is because for any collection of $(x,y)$ where $x,y $are in $\mathbb Q$, I can always find a point in $\mathbb R-  \mathbb Q$ which is a boundary point of that certain collection.  But is my thought ok ?  Would someone please help me on this?  I'm kinda lost >_&lt; Thanks ^^
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Prove that $\mathbb{R^2}\setminus\mathbb{Q}^2$ is connected.   First of all, I know this was     in this site.   However I'm looking for a proof that does not use path-connectedness (which I know, but this exercise is from a previous lecture), nor the knowledge that $\mathbb{R}^2\backslash E$, where $E$ is countable, is connected (it's the second part of the exercise).  I tried using disjoint open sets and separated sets, and also using the fact that they should also be the intersection of sets in $\mathbb{R}^2$ and $\mathbb{R}^2\setminus\mathbb{Q}^2$, but I hit a wall there.  I have the feeling I'm forgetting something very basic and obvious, or I may be overcomplicating things.
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Let $f:X\to Y$ be a homeomorphism between topological spaces. Suppose we have an equivalence relation $\sim$ defined on $X$. Define an equivalence $\sim'$ on $Y$ by $y_1\sim'y_2$ iff $x_1\sim x_2$ where $y_i=f(x_i)$     Is $X/{\sim}$ homeomorphic to $Y/{\sim'}$?   I have a well defined continuous map $F:X/{\sim}\to Y/{\sim'}$ given by $[x]\mapsto[f(x)]$. This can easily be seen to be bijective. I am having trouble proving that this is open. I know there is some diagram chasing involved but I am confused.  We have the following commutative diagram - $\require{AMScd}$ \begin{CD}     X @&gt;f&gt;&gt; Y\\     @Vq_XVV\circlearrowright @VVq_YV\\     X/{\sim}@&gt;&gt;F&gt; Y/{\sim'} \end{CD}  If $U$ is open in $X/\sim$ then $q_X^{-1}(U)$ is open in $X$ and $f(q_X^{-1}(U))$ is open in $Y$. To prove that $F(U)$ is open in $Y/\sim'$ we need to show that $q_Y^{-1}(F(U))$ is open in $Y$. This will be true if $f(q_X^{-1}(U))=q_Y^{-1}(F(U))$. But I am having trouble proving this is equality. Any help will be appreciated.  Thank you.
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Let $Q$ be the following set of $\mathbb{Z}\times \mathbb{Z}$   \begin{align*} Q=\{(a,b)\in \mathbb{Z}\times \mathbb{Z} | b\neq 0\} \end{align*}   Define the relation $\sim$ on $Q$ as   \begin{align*} (a,b)\sim (c,d)\Leftrightarrow ad=bc \end{align*}   Prove that $\sim$ is a relation of equivalence, and give the class of equivalence of $[(2,3)]$. Furthermore the general order of equivalence of $[(a,b)]$. Try to give a description $Q/\sim$.   I've asked the same question (I deleted it one hour later) but it didn't look well when I didn't try to add what I tried. Here we go. I have proven that $\sim$ is a relation of equivalence by showin it's i) reflexive, ii) symmetric and iii) transitive.   i) $\forall (a,b)\in Q:(a,b) \sim (a,b)$.   $\forall (a,b)\in Q:(a, b) \sim (a, b) \Leftrightarrow ab \stackrel{\surd}= ab$  ii) $(a, b) \sim (c, d) \Rightarrow (c, d) \sim (a,b)$  $(a, b) \sim (c, d) \Leftrightarrow ad = bc$ and $(c, d) \sim (a, b) \Leftrightarrow  cb = da$   then $ad = bc \stackrel{\surd}\Rightarrow cb = da$.  iii) $[(a, b) \sim (c,d) \wedge (c, d) \sim (e, f)]  \Rightarrow (a, b) \sim (e, f)$  $(a, b)\sim (c, d) \Leftrightarrow ad = bc$, $(c, d) \sim (e, f) \Leftrightarrow cf = de$ and $(a, b) \sim (e, f) \Leftrightarrow af = be$  then $((ad = bc) \wedge (cf = de))\stackrel{\surd}\Rightarrow af = be$.  The class of eq. $[(2,3)]$ is the class, that contains the element $(2,3)$, so for every element $(a,b)$ in this class satisfies $(a,b) \sim (2,3) \Leftrightarrow 3a=2b$, that is \begin{equation} [(2,3)] = \lbrace (a,b) \in \mathbb{Z}\times\mathbb{Z} | b \neq 0 \wedge 3a = 2b \rbrace \end{equation}  So now I've shown what I tried to show. Is there something I'm missing or something I should have mentioned? I need some help to answer the rest. Thanks.
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I asked a similar question yesterday about well ordered sets, now I am having troubles with equivalence relations.   Could someone suggest an injection from a well known set of cardinality $2^{\aleph_{0}}$ to the set of all equivalence relations of $\mathbb{N}$?  Many thanks in advance!