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Oct
15
comment Completion of rational numbers via Cauchy sequences
@AbstractionOfMe maybe it should be called "positivity of addition" or something. It's been so long I don't really want to edit this any more.
Oct
10
comment Closed Ball Complete iff $(M,d)$ is complete
@SujaanKunalan it seems you're not entirely clear on what the forward implication says. You're asked to prove that if $(M,d)$ is complete, then all closed balls in $M$ are complete (w.r.t. the induced metric, of course). But as I noted in my comment above, that's almost trivial. Concluding that $(M,d)$ is compact is irrelevant (and almost certainly false).
Oct
10
answered Closed Ball Complete iff $(M,d)$ is complete
Oct
10
comment Closed Ball Complete iff $(M,d)$ is complete
The forward implication can be proven much easier: a closed ball in a metric space has open complement and a closed subset of a complete space is complete.
Oct
10
answered Example of an infinitely differentiable function f : R → R with f(x) = 0 iff x = 0 and f intersects origin with infinite multiplicity
Sep
30
awarded  Explainer
Sep
15
comment Vector space without a scalar product
It's not true that an arbitrary vector space admits an inner product. It has to be a real or complex vector space. Now, this doesn't necessarily stop you from producing a non-degenerate symmetric bilinear form on your vector space (but it would not necessarily be an inner product if the vector space is complex as complex inner products have to be sesquilinear).
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