Thomas E.
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 Apr20 comment Maesurability of function $\mathbb R\ni t \mapsto P(X>t)$ what have you tried? 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 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 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 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. Mar24 comment $E$ compact, real-valued $f : E \to \mathbb{R}$ continuous iff graph is compact - is real valued necessary? @SantiagoCanez. You're right. $g^{-1}$ is clearly continuous as it's the restriction of a projection map on the first coordinate, but this detail is essential in order to conclude the continuity of $g$ given the other assumptions. Mar24 comment infinite Union of compact sets @pratik: Each $I_{n}$ is compact. But is $\mathbb{R}$ compact? Mar24 comment $E$ compact, real-valued $f : E \to \mathbb{R}$ continuous iff graph is compact - is real valued necessary? Can you elaborate why in the second direction of your proof: $g$ is continuous knowing that $G$ is compact and $g$ is bijective? This doesn't seem right. Mar19 comment Compact Projections to $S^1$ $\mathbb{R}^{2}$ is certainly not compact. Mar12 comment what can you conclude about the number of solutions of the linear system Ax = b? Why would $A$ have a right inverse? It could be that $a_{1},...,a_{4}$ span something lower dimensional that $\mathbb{R}^{3}$. In fact $A$ could be anything from the null matrix to full rank matrix here. Mar12 comment If $\| f \|_p \leq 1$, then $|f(x)| \leq 1$ for some $x$ Yeah, that would do the job. Mar12 comment what can you conclude about the number of solutions of the linear system Ax = b? @sovon: Which part did you not understand? Mar12 comment If $\| f \|_p \leq 1$, then $|f(x)| \leq 1$ for some $x$ @kobe: If $f>g$ then in general $\int f\geq \int g$. The pointwise consideration is not taking in to account any properties of the measure and the integration depends on the measure. I guess his question is how do you conclude here that the inequality is strict with the integrals? Might be trivial in the case of Lebesgue measure, but what is the argument?