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Doing math.


May
13
comment Geometric meaning of symmetric connection
Have you computed both sides of $(2)$ for the Euclidean connection on $\mathbb{R}^{n}$?
May
13
revised Motivation of vector bundle of a manifold
edited tags
May
13
answered Motivation of vector bundle of a manifold
May
11
comment Equivalent definitions of vector field
@soporhs. You're welcome. I'll gladly answer any questions you might have, but I advice that you try to work the details on your own at first.
May
11
revised Equivalent definitions of vector field
added 84 characters in body
May
11
comment Equivalent definitions of vector field
What definition of $TM$ are you working with? It seems that this answer only opens the usual definition of the tangent space but I can't see how you identify those two definitions of a vector field.
May
11
answered Equivalent definitions of vector field
May
11
comment Equivalent definitions of vector field
What is your definition of $TM$?
May
6
comment $\mathbb R$ has the same cardinality of any interval
$g$ is not defined at $x=-1$.
May
6
answered Openess of sets given by equivalence relations in the quotient topology.
May
2
comment $L^p$ not sequentially compact
Didn't you just show there that $\lim_{n\to\infty}\sqrt{sin(n\pi x)}= 0$ in the $L^{2}$ topology? Every subsequence of this sequence converges in $L^{2}$, so how is this a counter example?
May
2
answered Closure open ball
May
1
comment Show a bijection between sets
What are your thoughts so far? Have you tried some candidates?
Apr
29
answered Prove that $S=\left \{ (x,y)\in \mathbb{R}^2:x>0\, \; or \; y>0\left. \right \} \right. \subset \mathbb{R}^2 $is open
Apr
28
comment Is $A = f\mathbb{R^2}$ complete? $f:\mathbb{R^2}\to\mathbb{R^3}$, $f(x,y) = (x,y,x^2-y)$
@souf: Note that $f\mathbb{R}^{2}=\mathrm{graph}(g)$, where $g:\mathbb{R}^{2}\to \mathbb{R}$ as Goos defined above. So you only need to show that the graph of $g$ is a closed subset of $\mathbb{R}^{3}$ because closed subsets of complete spaces are complete.
Apr
28
answered Interior and closure of an arbitrary product
Apr
27
revised If $f$ is uniformly continuous on $\mathbb{R}$, $f(x) \ge a >0$ and $g(x) = 1/f(x)^2$, then $g(x)$ is uniformly continuous
edited tags
Apr
13
comment Why do some series converge and others diverge?
@Ethan: Because in the sum you add those numbers together. If you add numbers that are large then obviously the end result will be large. If you add negligent numbers together, then end result will stay small.
Apr
13
comment Why do some series converge and others diverge?
@Ethan: I mean that the sum is always less than $1$, for any finitely many terms that you sum together. By converging fast enough to zero I mean the following. By looking at the sequences $(\frac{1}{n})_{n=1}^{\infty}$ and $(\frac{1}{2^{n}})_{n=1}^{\infty}$ you will see that both of them converge to zero, but e.g. the third term in the first sequence is $\frac{1}{3}$, whereas in the second sequence it is $\frac{1}{8}$: much less. And for any given $n\in\mathbb{N}$‚ the other sequence will always have small values than the other. The terms go faster to zero so to say.
Apr
13
comment Why do some series converge and others diverge?
@Ethan: By a simple observation you can see that whenever you add finitely many of those terms $\frac{1}{2},\frac{1}{4},\frac{1}{8},...$ the sum is always bounded above by $1$. So the infinite sum must also be bounded above by $1$. The infinite sum is defined as the limit of the partial sums. they might be diverging or converging to a specific number, that all depends on the numbers that you're adding. Usually you can think that if the terms of the sum converge "fast enough" to zero, then the corresponding sum is finite.