58,911 reputation
53798
bio website
location
age
visits member for 2 years, 4 months
seen 3 hours ago

17h
comment Show that the set of all finite subsets of $\mathbb{N}$ is countable.
@Mark: Not that it's probably relevant to the OP, but a countable union of finite sets needn't be countable in general without sufficient Choice. Here, of course, we're dealing with well-ordered finite sets, so it works out.
Jul
23
revised Are there real-life relations which are symmetric and reflexive but not transitive?
added 75 characters in body
Jul
19
revised How can I show that the closure of the set $\Phi = \{1, 2, 3, 4,\ldots\}$ is the set itself?
added 1 character in body
Jul
7
comment Alternatives to quotient rule
Oops! That's what I get for trying to do this while half-asleep. That should be $$\frac{5x-5a+5a^2x-5ax^2}{(1+x^2)(1+a^2)(x-a)},$$ instead.
Jul
7
comment Alternatives to quotient rule
You're almost there in your second-to-last step! Instead, rearrange it as $$\frac{5x-5a+5a^2x-5ax^2}{x-a},$$ and try factoring the top by grouping from there. See if that gets you the rest of the way, but let me know if you get stuck again.
Jul
7
revised Alternatives to quotient rule
added 42 characters in body
Jul
7
answered Alternatives to quotient rule
Jul
6
comment My proof is wrong, can anyone tell me why?
Oliver: I believe what @André is trying to say is that we must restrict which pairs of integers we're looking at. In particular, we need to look at $x,y\in\Bbb Z$ such that $x+y\neq-1.$ Given that restriction, we will have $x(x+1)=y(y+1)$ if and only if $x=y.$
Jul
6
revised My proof is wrong, can anyone tell me why?
added 271 characters in body
Jul
6
comment My proof is wrong, can anyone tell me why?
If you want to disprove a universal statement, then all you have to do is find one counterexample. If you wish to disprove the universal statement $$\forall x\in\Bbb Z,\forall y\in\Bbb Z,[x^2-y^2=0]\iff[x-y=0],$$ for example, then you need to find some $x,y\in\Bbb Z$ for which the biconditional $$[x^2-y^2=0]\iff[x-y=0]$$ does not hold. One of the conditionals will always hold, so you'd need to find $x,y\in\Bbb Z$ such that $x^2-y^2=0$ and $x-y\neq0.$
Jul
6
revised My proof is wrong, can anyone tell me why?
added 500 characters in body
Jul
6
answered My proof is wrong, can anyone tell me why?
Jul
2
comment Basic calculation with root roots and power
You're very welcome! (Apologies for not responding sooner.)
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jul
2
comment increasing sequence in $\omega_1$
It certainly does. Let's call the subset $S.$ To confirm it as an initial segment of $\omega_1,$ you need only show that if $\alpha,\beta\in\omega_1$ with $\alpha<\beta$ and $\beta\in S,$ then $\alpha\in S.$ In the case that $\beta\neq b,$ this is fairly trivial, and not too much harder when $\beta=b.$
Jul
1
comment increasing sequence in $\omega_1$
The cofinality of $\omega_1$ is equal to $\omega$ if and only if $\omega_1$ is a countable union of countable sets, no?
Jul
1
revised increasing sequence in $\omega_1$
added 115 characters in body
Jul
1
comment increasing sequence in $\omega_1$
Apologies for the first version of the answer. It was patently incorrect, in addition to being overly advanced. Please see my corrected and improved answer, and let me know if you've any questions.
Jul
1
revised increasing sequence in $\omega_1$
added 700 characters in body