7,867 reputation
11437
bio website
location
age 42
visits member for 1 year, 9 months
seen 7 hours ago

8h
reviewed Approve suggested edit on how does this converges? Sequence and series convergence
8h
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. :-)
9h
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.
9h
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?
1d
awarded  Good Answer
1d
awarded  Nice Answer
1d
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 $1=0$.
1d
answered Is it possible that “A counter-example exists but it cannot be found”
1d
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.
1d
comment Subsets of $[0,1]$
Note that you have to intersect with $(0,1)$, not $[0,1]$ because an open set was requested.
1d
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
revised How do you sketch a parabola given the equation?
LaTeX for formula
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 $\mathbb R$.