Reputation
4,110
Next privilege 5,000 Rep.
Approve tag wiki edits
Badges
2 11 30
Impact
~54k people reached

Apr
9
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
Apr
9
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 $V$?
Mar
29
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.
Mar
27
revised Why the column space of a matrix is useful?
added 393 characters in body
Mar
27
answered Why the column space of a matrix is useful?
Mar
27
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?
Mar
27
answered Integrals of indicator functions question
Mar
27
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?
Mar
27
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.
Mar
27
comment Define Radon measure as an integral
What is the definition of an outer Radon measure that you're working with?
Mar
27
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.
Mar
26
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.
Mar
26
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
Mar
26
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.
Mar
26
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.
Mar
26
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 $x$?
Mar
26
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 $a_2 =(-1,3)$.
Mar
26
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.
Mar
26
answered Prove $f$ is diagonalizable iff $V=W \oplus Z$ where $W,Z \subseteq V$ are $f$ invariant
Mar
24
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.