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Sep
11
comment Is the sum of the diagonals of an isosceles trapezoid at least the sum of the bases?
@Sigur: Do you mean the triangle inequality? I hadn't so far, but if $a,b$ are bases and $c$ the leg, $d$ the diagonal, then triangle inequality would give $a+b\leq c+d+c+d=2d+2c$. The $2c$ kind of ruins it:( Or did I misunderstand your point?
Sep
11
asked Is the sum of the diagonals of an isosceles trapezoid at least the sum of the bases?
Aug
31
comment How to construct a one-to one correspondence between$\left [ 0,1 \right ]\bigcup \left [ 2,3 \right ]\bigcup ..$ and $\left [ 0,1 \right ]$
@AsafKaragila: Just out of curiosity, what argument did you have in mind that would contradict continuity from this union set to $[0,1]$?
Aug
27
comment How to strictly mathematically prove that definition is wrong?
What do you mean by proving that $5=4$? What is $5$?
Aug
27
answered Does a compact subspace need to be closed?
Aug
24
comment the function $1/|x|^α$ in $R^n$
@BR. Thanks, that makes it a lot more clear.
Aug
24
comment the function $1/|x|^α$ in $R^n$
If $f:\mathbb{R}^{n}\to\mathbb{R}$, then what does $f(r)$ mean for $r\in[0,\infty[$?
Aug
23
answered Question about measurable sets $E_n$ such that $\lim_{n\rightarrow \infty}L^N(E_n) = 0$.
Aug
17
comment 2 jars with 50 balls each. Pick any ball with $>50\%$ probability.
@sTEAK. If you take one from each jar then the probability of getting a red is $1$.
Aug
16
comment Is it possible to have $D=\Bbb P$
@t.b. Yeah, the countability is important :) Baire implies that any countable completely metrizable topological space has an isolated point. If not, you could write the whole space as a countable union of nowhere dense closed sets (singletons), making the whole space to have an empty interior, which is a contradiction.
Aug
16
comment Is it possible to have $D=\Bbb P$
@t.b. Yeah, I meant an isolated point. And true, that also does the job. Thanks for providing an alternative way of seeing it.
Aug
16
revised Is it possible to have $D=\Bbb P$
added 462 characters in body
Aug
16
answered Is it possible to have $D=\Bbb P$
Aug
14
comment Is the set of irrationals separable as a subspace of the real line?
@CameronBuie. To add on Asaf's comment, it's worth noting that the word 'metric space' plays a key role there. Metrizability is hereditary and so is second countability. Since in a metric space separability (also Lindelöf) is equivalent with second countability, then every subspace of a separable metric space is separable (and Lindelöf, for that matter).
Aug
13
comment Baire Category Theorem
@AsafKaragila. True. The ZF part was so well hidden in the link that I totally missed it.
Aug
13
comment Baire Category Theorem
By the way, this works also for non-separable spaces. And you could try to show the complement of this statement which says that a countable intersection of open dense sets is dense. It's quite simple, too. You begin with an arbitrary open ball and start choosing closed balls inside it inductively so that the radius goes to zero and the $n$:th step ball is a subset of $n$ first members of the dense open sets. The intersection of these closed balls is a singleton by completeness: it belongs to the intersection of the open dense sets and our original ball, proving the claim.
Aug
13
comment $L^p$-norm of a non-negative measurable function
Hint: Assume first that $f$ is a simple function and show that the equation holds. Then for your non-negative measurable $f$ choose a nondecreasing sequence of simple functions converging point-wise to $f$ and use monotone convergence theorem.
Aug
13
comment Outer measure of a union of 2 subsets of disjoint measurable sets of real numbers.
Why are subsets of measurable sets bounded? Consider for example $]0,\infty[\subset \mathbb{R}$, where $\mathbb{R}$ is measurable and $]0,\infty[$ is unbounded.
Aug
6
comment How to `bound' $L^\infty$ by the constant function $1$
@t.b. Thanks. I overlooked that part and it makes perfect sense now.
Aug
6
comment How to `bound' $L^\infty$ by the constant function $1$
But this bound only works almost everywhere. Sure, you can find a representative of $f$ from same equivalence class for which this works, but in general $f$ can be unbounded. Right?