There's the one story of a bet he had with a friend of his that involved whether he could quit amphetamines for a month. He did, but said something to the effect of that being without them hampered his mathematical creativity
There is the story where PE tried to open a box(?) of tomato-juice with a knife (he never did these things himself, even could't cut bread to slices or smear butter on it, it did all his mom, with whom he lived.) So Graham found him in the middle of the night in his kitchen all over with a deep red liquid an a knife in his hand ...
I am looking for a way to create four-dimensional plots (surface plus a color scale) using Python and matplotlib. I am able to generate the surface using the first three variables, but I am not having success adding the color scale for the fourth variable. Here is a small subset of my data belo...
Let $\{v_1,\ldots,v_n\}$ be vectors in a vector space. One way to show that they are linearly independent is to show that each $v_i$ cannot be written as a linear combination of the others. If I remember correctly, this algorithm can be slightly reduced. Does it suffice to show that for each $i \in \{1,\ldots,n\}$ the vector $v_i$ cannot be written as a linear combination of the $v_j$ with $j < i$ ?
Subspace generated by v_1 has dimension less than subspace generated by v_1 and v_2, etc. So the dimension of the subspace spanned by $v_1,\dots,v_n$ is at least $n$.
@abenthy Sure, it's true and I've seen it. Another more general version for direct sums appears in Isaac's algebra book, for example.
@abenthy Think of it this way: if the whole set is linearly dependent, you can pick the highest indexed basis element appearing in a nontrivial linear combination, and then that's written in terms of basis elements of lesser index.
Hi all, revival of my yesterday confusion about limits, I think I got it anyway but just for cross checking: $$0\in\lim_{n\to\infty}\bigcup_{m\in\{1,\dots,n\}} \big(\frac{1}{m}\big)$$ with $(a_n)$ being the sequence of $a_n$.
Let $\mathcal{H}$ be some Hilbert space, let $B(\mathcal{H})$ denote the bounded linear operators acting on $\mathcal{H}$, and let $M_n(B(\mathcal{H}))$ denote the $n \times n$ matrices with operator-valued entries. Let $A = [a_{ij}]$ be one such matrix. My question is,
If $\sum_{i,j=1}^n u_i...
@Rudi_Birnbaum First of all, I can easily imagine looking at the limit of the sequence $\frac{1}{m}$ and concluding it is $0$. You're sure that isn't the point?
@Rudi_Birnbaum It looks like you want to use (1/n) for both a one element set and the set of all such reciprocals? You're going to have to pick, I'm afraid.
If $A=\{\frac{1}{n}\mid n\in \mathbb N^+\}$, then yes, the closure in $\Bbb R$ does happen to be $A\cup\{0\}$. $A$ is a subset of $\Bbb R$ missing point of closure, $0$.
Is it correct that the set must have the structure of a topological space in order to be able to define a limit? Or is there no connection with a limit construction at all?
@Rudi_Birnbaum Not necessarily. In the case of the ordinary topology on the reals, the limit of that sequence happens to be the only missing point of closure
Yeah we were digressing, but my question is answered! Thank you very much!
Maybe to summarize: I have ignored the definition of union and found out that direct "generalization" of a limit for unions does not exist. But there exists a construction called closure for sets of elements of topological spaces and which however also has some common grounds with limit constrcutions. OK?
That's ok. I understand, @MikeM. Happy to respond to questions if there are any. Your question is way more straightforward (because of the unique normal direction) than it would be in higher codimension. It was just one paragraph on the second page.
@rschwieb Sorry just one more question to be sure I git that about the union: Is $\bigcup_{n\in\Bbb N^+} (1-\frac{1}{n},1-\frac{1}{n+1}] = (0,1)$ correct? with $(a,b]$ the half open interval from without $a$ to including $b$.
@rschwieb Yeah somehow here I can easily understand why $1$ is not contained, while it still feels weakly odd that its not in $(1/n)_{n\in\Bbb N^+}$. But I see now the necessity for this very definition of union.
for intervals it makes more sense somehow (in my eyes)..
the example shows that the arbitrary union of closed sets doesn't have to be closed.
Let $\mathcal{H}$ be some Hilbert space, let $B(\mathcal{H})$ denote the bounded linear operators acting on $\mathcal{H}$, and let $M_n(B(\mathcal{H}))$ denote the $n \times n$ matrices with operator-valued entries. Let $A = [a_{ij}]$ be one such matrix. My question is,
If $\sum_{i,j=1}^n u_i...
