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Oct
14
comment Understanding the quotient of infinite groups $\mathbb{R}^2/H$ where $H = \{(a, 0): a\in \mathbb{R}\}$
@blehdsf: I wouldn’t index them initially: I’d just talk about $\{\langle x,y\rangle+H:\langle x,y\rangle\in\Bbb R^2\}$. Once I’d proved that $\langle x,y\rangle+H=\langle u,v\rangle+H$ iff $y=v$, then I could index them as $\{C_y:y\in\Bbb R\}$, where $C_y=\langle 0,y\rangle+H$, say.
Oct
14
answered Understanding the quotient of infinite groups $\mathbb{R}^2/H$ where $H = \{(a, 0): a\in \mathbb{R}\}$
Oct
14
comment Defining a recursive function $f$ on $\{a, b\}$*
@Dabbish: Suppose that $w=abb$, so that $f(w)=baa$; according to your rule $f(abba)=abaa$; is that actually correct? // You do not want to consider $f(bwa)$ or $f(awb)$. Every word in $\{a,b\}^*$ can be built from shorter ones simply by appending $a$ or $b$, and the definition of this particular function does not require looking at anything more complicated than $wa$ and $wb$.
Oct
13
comment Defining a recursive function $f$ on $\{a, b\}$*
@Dabbish: That rule would apply only to words beginning with $a$ and ending with $b$, so it could not, for instance, tell you what $f(b)$ and $f(ba)$ are, and it definitely would not do what you want even when it does apply to an input: $f(awb)$ must certainly begin with $b$, not $a$. Once again, if you know what $f(w)$ is, how can you find $f(wa)$?
Oct
13
answered Defining a recursive function $f$ on $\{a, b\}$*
Oct
13
comment How can I solve this recursion function task?
@Dabbish: You’re welcome.
Oct
13
comment How can I solve this recursion function task?
@Dabbish: Yes, and no, in that order. You want $r(w\alpha)=\alpha r(w)$ for any word $w$ and letter $\alpha$. The idea is to say how to construct the reversal of the longer word if you already know the reversal of the shorter one.
Oct
13
comment Prove that R∪S is an equivalence relation on A∪B. A and B are disjoint.
The disjointness of $A$ and $B$ isn’t a hindrance: it’s needed in order to show that $R\cup S$ is transitive.
Oct
13
comment How can I solve this recursion function task?
@Dabbish: Let’s try once more; if you don’t get it from this question, I’ll simply give you the answer. When you add a letter to the righthand end of $w$, what happens to the reversal? You add that same letter ... where?
Oct
13
revised Homeomorphism preserving partitions
added 459 characters in body
Oct
13
comment Homeomorphism preserving partitions
@ChaoXu: That’s not enough. Let $X=Y=[0,2)\cup [3,4]\cup[5,6]$, $X_1=[0,1)\cup[3,4]=Y_2$, and $X_2=[1,2)\cup[5,6]=Y_1$. The map from $X$ to $Y$ that interchanges $[3,4]$ and $[5,6]$ meets your requirement, but there’s still no homeomorphism that interchanges $X_1$ and $Y_1$.
Oct
13
comment Please help me interpreting Elementary Set Theory question
@Kurow: You’re welcome!
Oct
13
answered Homeomorphism preserving partitions
Oct
13
comment Proving $N(A ∪ B) = N(A) + N(B) − N(A ∩ B)$
... $$N\big(\big(A\setminus(A\cap B)\big)\cup B\big)=N\big(A\setminus(A\cap B)\big)+N(B)\;.$$
Oct
13
comment Proving $N(A ∪ B) = N(A) + N(B) − N(A ∩ B)$
@Eric: You may have the right idea in mind, but what you’ve said isn’t right. $A\cap B$ may be non-empty, so $A$ and $B$ need not be disjoint. However, $A\setminus(A\cap B)$ and $B$ are disjoint: if $x\in A\setminus(A\cap B)$, then $x\in A$ and $x\notin A\cap B$. Since $x\notin A\cap B$, either $x\notin A$ or $x\notin B$ (or both), but we know that $x\in A$, so $x\notin B$. Thus, $x\in A\setminus(A\cap B)$ implies that $x\notin B$, and it follows that $A\setminus(A\cap B)$ and $B$ are disjoint, and hence that ...
Oct
13
answered Please help me interpreting Elementary Set Theory question
Oct
13
comment Proving $N(A ∪ B) = N(A) + N(B) − N(A ∩ B)$
@Eric: $A$ is a set of objects; $N(A)$ is the number of objects in that set. You use $A$ when you want to talk about the set; you use $N(A)$ when you want to talk about the number of objects in that set. Thus, it makes no sense to write $A+B$: $A$ and $B$ are sets, and addition is not an operation on sets. It does make sense, however, to write $N(A)+N(B)$: $N(A)$ and $N(B)$ are integers, and we do have an operation of addition on the integers. // $A\cup B$ is not necessarily a disjoint union, but $\big(A\setminus(A\cap B)\big)\cup B$ is one; why?
Oct
13
comment Proving $N(A ∪ B) = N(A) + N(B) − N(A ∩ B)$
@Eric: Well, you need some $N$’s in there, but yes, that’s the idea.
Oct
13
comment Proving $N(A ∪ B) = N(A) + N(B) − N(A ∩ B)$
@Eric: Yes, that’s exactly right. And I’m fairly sure that you’re not expected to prove those two facts, unless your course has dealt pretty rigorously with the notion of finite cardinalities. Can you see how to use those facts to calculate $N(A\setminus(A\cap B)$ and then the desired result?
Oct
13
answered Help with Induction problem?