Thomas E.
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 Apr20 comment Maesurability of function $\mathbb R\ni t \mapsto P(X>t)$ what have you tried? Apr9 revised Why does closedness and boundedness for $S = \{ v \in V : || v||_{\infty} = 1\}$ imply that $S$ is compact in a finite dimensional vector space $V$? added 1 character in body Apr9 answered Why does closedness and boundedness for $S = \{ v \in V : || v||_{\infty} = 1\}$ imply that $S$ is compact in a finite dimensional vector space $V$? Mar29 comment Let $X$ be a topological space. Prove that for any $x$ in the intersection of all opens sets $=\{x \}$, the space $X$ need not be Hausdorff. This is answering a different question than what you formulated in the problem statement in the comments. Saying that the intersection of all open sets around a point will yield the singleton, is not the same as saying that any two open sets containing a point will yield the singleton. With Zariski topology the intersection of two open sets is always uncountably infinite, while with uncountably infinite intersections you can get the singleton. Mar27 revised Why the column space of a matrix is useful? added 393 characters in body Mar27 answered Why the column space of a matrix is useful? Mar27 comment A problem in Weierstrass M-test proof For a sequence $(a_{n})_{n=1}^{\infty}$, if $|\sum_{k=1}^{N}a_{k}|\leq \sum_{k=1}^{N}|a_{k}|$ for all $N\in\mathbb{N}$ by the triangle inequality, then taking limits on both sides you get $|\sum_{k=1}^{\infty}a_{k}|\leq \sum_{k=1}^{\infty}|a_{k}|$, if both limits exist. So what is the problem of using triangle inequality infinitely many times? Mar27 answered Integrals of indicator functions question Mar27 comment finding formula for M^n What exactly is the matrix $M$ here? Are those the column or the row vectors of $M$ that you have given there? Mar27 comment Proof by contradiction to prove an inequality does not hold If you want to show that there is no positive integer such that ... , the counter assumption is that there is such a positive integer. Not, that it is true for all positive integers. As for a hint how to proceed: as Andre Nicolas said you can just divide by $x$ to get a contradiction. Mar27 comment Define Radon measure as an integral What is the definition of an outer Radon measure that you're working with? Mar27 comment Let $X$ be a topological space. Prove that for any $x$ in the intersection of all opens sets $=\{x \}$, the space $X$ need not be Hausdorff. @Zeta10: Well the Zariski topology on $\mathbb{R}$ is obviously not going to work, because every open set has a finite complement. So any two open sets have an uncountable intersection. Mar26 comment Let $X$ be a topological space. Prove that for any $x$ in the intersection of all opens sets $=\{x \}$, the space $X$ need not be Hausdorff. @Zeta10: Yes. Correct. Mar26 revised Let $X$ be a topological space. Prove that for any $x$ in the intersection of all opens sets $=\{x \}$, the space $X$ need not be Hausdorff. edited body Mar26 answered Let $X$ be a topological space. Prove that for any $x$ in the intersection of all opens sets $=\{x \}$, the space $X$ need not be Hausdorff. Mar26 comment Let $X$ be a topological space. Prove that for any $x$ in the intersection of all opens sets $=\{x \}$, the space $X$ need not be Hausdorff. Ok. So the assumption is that if two open sets intersect, then the intersection is a singleton. Mar26 comment Let $X$ be a topological space. Prove that for any $x$ in the intersection of all opens sets $=\{x \}$, the space $X$ need not be Hausdorff. Can you rephrase the assumption? Do you mean that if $x$ is in the intersection of two open sets then this intersection must be equal to the singleton of $x$? Mar26 comment Showing a matrix is not diagonalizable @NajibIdrissi: You have it the wrong way around. There are normal matrices that are not diagonalizable: but certainly all diagonalizable matrices are normal, by looking at the diagonal decomposition you should see this. For a normal non-diagonalizable matrix take $M=(a_1 , a_2)$ where the column vectors are $a_1 = (1,1)$ and $a_2 =(-1,3)$. Mar26 comment Showing a matrix is not diagonalizable In your second paragraph you probably meant to say that the geometric multiplicity is strictly smaller than its algebraic multiplicity, instead of repeating the word geometric. Mar26 answered Prove $f$ is diagonalizable iff $V=W \oplus Z$ where $W,Z \subseteq V$ are $f$ invariant