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seen Jan 11 at 19:47

Apr
17
comment Is there a more concise way?
@naslundx: Well, that must be the reason why Fermat could not write it on the border of the page. :-)
Apr
17
comment Is it possible that “A counter-example exists but it cannot be found”
Thinking again about my parenthetical comment in the previous comment: The well-ordering might itself be impossible to specify, therefore this trick probably doesn't work. Anyway, the fact that the very same argument can be applied to the ordinals and gives a contradiction already shows that the argument as such cannot be correct.
Apr
17
comment Is it possible that “A counter-example exists but it cannot be found”
Interesting. I can't say I understand all of what was discussed, but the example with the ordinals in the linked MathOverflow answer is quite convincing that something is wrong with the argument (@EricTowers: There exist nonempty sets of positive real numbers without a minimum — most notably the set of positive real numbers itself —, so this is no argument; however one could impose a well-ordering on the reals by mapping to an ordinal of the right cardinality and make the argument with that ordering). I'm not sure what I now should do with my answer. Just delete it? Amend it saying it's wrong?
Apr
17
awarded  Good Answer
Apr
16
awarded  Nice Answer
Apr
16
comment Find the solutions of: $\sin x+\cos x=\sin^2 x+0.5\sin{2x}$
The equation as written is wrong: Inserting $x=0$ gives $1=0$.
Apr
16
answered Is it possible that “A counter-example exists but it cannot be found”
Apr
16
comment Subsets of $[0,1]$
OK, I interpreted it as subset of the $[0,1]$ interval in $\mathbb R$ (with the topology of $\mathbb R$). Anyway, when intersecting with the open interval, it's true for both.
Apr
16
comment Subsets of $[0,1]$
Note that you have to intersect with $(0,1)$, not $[0,1]$ because an open set was requested.
Apr
16
comment Subsets of $[0,1]$
It will not have the measure $\epsilon$ due to overlapping intervals (think of the rationals which lie inside the very first interval, for example). It will, however, have measure $\mu$ with $0<\mu<\epsilon$. In addition, you have to intersect with $(0,1)$ because otherwise it may not be a subset of $[0,1]$.
Apr
14
comment Probability of at least two events occurring.
If the question was asked at a meeting of the yellow party, the probability is negligible. ;-)
Apr
14
answered A question in combinatorics
Apr
13
revised If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not?
Added a short initial description about what the proof does
Apr
13
answered If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not?
Apr
13
comment If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not?
Well, showing that is showing that $\mathbb R$ is connected.
Apr
13
comment If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not?
With empty border. You prove the original statement by proving that.
Apr
13
comment If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not?
More exactly, its border is the empty set (the border as set of all border points is always defined, but the clopen set has no border points, and thus the border is empty).
Apr
13
comment If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not?
The border obviously cannot contain elements outside of $\mathbb R$ because we are talking about the topology of $\mathbb R$.
Apr
13
comment If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not?
A closed set contains all of its border. An open set contains none of its border. What do you conclude about the border of a set that is both open and closed?
Apr
13
comment Finding the largest factor of a number, possible without exhaustion?
Indeed, whoever finds an efficient algorithm will get famous. Unless an intelligence agency decides it's better to prevent him from publishing it ...