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seen May 30 at 19:50

Jul
2
awarded  Curious
May
18
comment A problem with concyclic points on $\mathbb{R}^2$
Awesome proof. And since it depends heavily on Sylverster-Gallai it is very unlikely a naive proof as I attempted would work.
May
18
comment A problem with concyclic points on $\mathbb{R}^2$
Maybe I misunderstand, but in Lemma 1, you take $C'$ to be the intersection points between lines $Q_1Q_j$ with $\Pi$, and not the projections of $Q_j's$ onto $\Pi$, right?
May
18
comment A problem with concyclic points on $\mathbb{R}^2$
I posted it as an answer, way too long fr a comment.
May
18
answered A problem with concyclic points on $\mathbb{R}^2$
May
17
accepted Operators with finite spectrum
May
17
comment A problem with concyclic points on $\mathbb{R}^2$
Very interesting problem! I did it for 4,5,6,7,and 9 points, it is a fairly easy counting argument. But other than that....I tried to do it for 8 points, but I could not.
May
15
accepted Density of union of closed sets
May
15
comment Density of union of closed sets
I see, I think. If you take $V_n$ a multiple of the closed unit ball of $[x_1,x_2,\dots x_n]$ such that it contains all $x_k's$, for $1\leq k\leq n$, and also contains the previous $V_{n-1}$. Each $V_n$ has empty interior, so by Baire, their union cannot be $X$. Does this look right?
May
14
asked Density of union of closed sets
Apr
25
awarded  Yearling
Apr
24
answered Can a cube always be fitted into the projection of a cube?
Apr
21
comment Subsequences of a basic sequence
Oh, it says basic sequence, not basis, I read it wrong. I think the question is nontrivial if $(x_n)$ is a basis, or if $Y$ is infinite codimensional in the closed span of $(x_n)$. @HowAboutaNiceBigCupof
Apr
20
awarded  Investor
Apr
20
comment Subsequences of a basic sequence
That's an interesting question. Definitely true for finite dimensional $Y$.
Apr
17
revised Show that $\lambda \in \sigma(A),$ $\lambda$ not an eigenvalue, implies that $\lambda \in \sigma(A + K)$ where $K$ is compact.
added 5 characters in body
Apr
17
revised Show that $\lambda \in \sigma(A),$ $\lambda$ not an eigenvalue, implies that $\lambda \in \sigma(A + K)$ where $K$ is compact.
edited body
Apr
17
answered Show that $\lambda \in \sigma(A),$ $\lambda$ not an eigenvalue, implies that $\lambda \in \sigma(A + K)$ where $K$ is compact.
Feb
26
comment Behavior of the resolvent near the boundary of the spectrum
Isn't the above true regardless of whether $T$ has eigenvalues? $(\lambda I-T)^{-1}x$ cannot have an entire extension, therefore some isolated singularity $\mu$ is either essential or pole. Either case gives that $(\lambda I-T)^{-1}x$ is unbounded near $\mu$. This should work in the case when spectrum is a convergent sequence (the limit becomes isolated if all other singularities can be removed), or, even more generally, when we can remove all isolated singularities in finite number of steps. Hope this makes sense. Thank you for your counterexample.
Feb
25
accepted Behavior of the resolvent near the boundary of the spectrum