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Aug
17
awarded  Yearling
Aug
15
comment How to get a function similar to the ELO ranking system
It's "Elo", not "ELO". It's not an abbreviation. It even says as much in the Wikipedia article you linked.
Aug
2
reviewed Reject suggested edit on integral of $\frac{1}{(1+e^{-x})}$
Jul
29
comment Why is this directed graph strongly connected?
There are no arrows on the (1,4) edge. So you can go in both directions there?
Jul
15
comment What if we removed all the irrational numbers from the real number line?
You might find this enlightening: dpmms.cam.ac.uk/~wtg10/mathsindex.html. Especially the section on analysis.
Jul
10
revised Span of Dirac's delta distributions dense in Hilbert space of $L^2$ functions?
Use \langle and \rangle (or their unicode equivalents) instead of < and >
Jul
9
comment The binomial formula and the value of 0^0
But the use of $\sum_{i} |x_i|^0$ as the definition of the zero "norm" seems to me to have basically nothing to do with exponentiation since it's actually computing the Hamming distance to $0$. And if one argues that it can be realised as the limit of $p$-norms as $p$ goes to zero, then continuity and limits are coming back in to the picture and it's no surprise that all bets are off.
Jul
8
comment Basis in the vector space of all polynomials
Furthermore, the Fredholm alternative doesn't apply in the case of $\mathbb R[x] \cong c_c$ since it's not complete.
Jul
7
comment Every uncountable subset of $\mathbb{R}$ has a limit point
Here's a much simpler argument: There must be an $n$ such that $A := E \cap [-n,n]$ is infinite. $A$ is an infinite subset of the compact set $[-n,n]$ so it has a limit point in $[-n,n]$. Now argue that such a limit point must also be a limit point of $E$.
Jul
2
comment Why do determinants have their particular form?
One thing that one has to verify regarding $T(\bigwedge_{i=1}^n v_i) := \bigwedge_{i=1}^n T(v_i)$ is that there could be different vectors $w_1,\dotsc,w_n$ such that $\bigwedge_{i=1}^n v_i = \bigwedge_{i=1}^n w_i$, so you have to check that you get the same result on the right hand side regardless of the choices you make.
Jul
2
comment Why do determinants have their particular form?
The multilinearity and alternatedness of the determinant suggests a connection with the exterior algebra of the vector space in question. And indeed there is such a connection (q.v. the linked article on Wikipedia).
Jul
2
awarded  Curious
Jun
29
comment Topology of convergence in measure
@Adam $L^0$ usually denotes the set of measurable functions. I don't think the topology it's usually given is normable (Like $L^p$ for $0 < p < 1$ isn't). en.wikipedia.org/wiki/Convergence_in_measure#Topology
Jun
23
comment Is the arbitrary union of consecutively non-disjoint path-connected sets path-connected?
@EricStucky the problem is that any point $(\alpha,a)$ has uncountably many copies of $[0,1)$ between it and $\infty$.
Jun
23
comment Is the arbitrary union of consecutively non-disjoint path-connected sets path-connected?
@EricStucky after some thinking it seems that the extended long line might not work. The point at infinity isn't connected by a path to anything else.
Jun
19
comment What are some motivating examples of exotic metrizable spaces
$B_1$ is also an excellent example of an open ball with many centres: $B_1 = B(a,1-a)$ for any $0 \leq a < \tfrac12$. Although it's not as extreme as the situation in an ultrametric space
Jun
18
revised Riesz Lemma for reflexive spaces
edited body
Jun
18
comment Let $T: \Bbb R^3 → \Bbb R^2 $be a linear transformation defined by $T( x, y, z) = ( x + y, x - z)$
LaTeX tips: Do not use \Bbb it's a mathjax thing and won't work in actual LaTeX files. Use \mathbb instead. E.g. \mathbb R: $\mathbb R$. You should use \dim to get $\dim$. It's useful to use abbreviations like \ker T ($\ker T$) and \operatorname{im} T ($\operatorname{im} T$). Yes, LaTeX and mathjax don't come with a predefined \im and \Im gives the symbol for imaginary part of a complex number.
Jun
16
revised Why in this case $f$ should be entire?
added 20 characters in body
Jun
14
answered Is there any false case for that: $\exists x \in D, \forall y \in D, P(x, y) \implies P(y, x)$?