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Apr
22
comment Reflexive for belonging ($\in$)
@Sibi: Because one could imagine that there are sets which contain themselves as elements, for example $A=\{A\}$. For that set, if it existed, $A\in A$ would be true.
Apr
22
comment Reflexive for belonging ($\in$)
@Sibi: $\in$ is "is element of". The set $\{0,1\}$ has two elements, namely $0$ and $1$. That is, $x\in A$ if and only if either $x=0$ or $x=1$.
Apr
22
comment Reflexive for belonging ($\in$)
Actually, in the standard construction, $0$ and $1$ are sets, namely the empty set, and the set containing (only) the empty set. Indeed in that construction, $A$ is also a number, namely the number $2$.
Apr
21
comment How many different messages can be transmitted in n microseconds using three different signals…
You can also start at $0$ by noting there's exactly one message you can send in $0$ microseconds, namely the empty message (no signal). Then the recursion naturally gives you $a_2=a_1+2a_0=1+2\cdot 1=3$.
Apr
21
comment Knot theory: Braids
Did you try drawing both braids?
Apr
21
comment If $A + B = \frac{\pi}{3} (A,B>0),$ Then the minimum value of sec A + sec B is?
Hint: $B=\pi/3-A$
Apr
21
revised Plotting a region in $3D$ space
added 74 characters in body
Apr
21
revised Plotting a region in $3D$ space
Added labels and image example
Apr
21
answered Plotting a region in $3D$ space
Apr
21
comment Plotting a region in $3D$ space
Anyway, I just now notice that you specified a region in 2D ("$(x,y)\in\mathbb R^2$") with a parameter $z$. Did you possibly mean "$(x,y,z)\in\mathbb R^3$"?
Apr
21
comment Plotting a region in $3D$ space
Mathematica should be able to do that (with RegionPlot3D). However given that you tagged your question with maple I'm not sure that this answers your actual question (although it answers the question you've written).
Apr
21
accepted Non-constructive axiom of infinity
Apr
21
comment Is it possible that “A counter-example exists but it cannot be found”
OK, I changed that. Now OK?
Apr
21
revised Is it possible that “A counter-example exists but it cannot be found”
added 32 characters in body
Apr
21
comment Non-constructive axiom of infinity
Yes, that's the contradicton I mentioned. But where it the induction?
Apr
21
comment Non-constructive axiom of infinity
Hmmm ... with $A=\omega\cup\{\{42\}\}$ I get $\bigcup A=\omega\cup\{42\}\ne A$.
Apr
21
comment Non-constructive axiom of infinity
OK, I think I see now why your proof works, but I don't see where you use induction. As far as I can see you just build a contradiction by proving $\rank(\bigcup x)<\rank x$, which contradicts $\bigcup x=x$.
Apr
21
comment Non-constructive axiom of infinity
Ah, OK, that makes more sense. Now let me get back to understanding the rest. :-)
Apr
21
comment Non-constructive axiom of infinity
"We can prove that there are only finitely many sets of finite rank" ... but don't all natural numbers in the standard construction have finite rank? There are definitely infinitely many natural numbers (although each of them is of course a finite set in the standard construction).
Apr
21
comment n-dimension hypercube!
Duplicate? math.stackexchange.com/questions/763458/…