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Doing math.


Jan
29
comment Showing $S^n$ is compact
By the way, note that Bounded + Closed $\Rightarrow$ Compact is not true in general, but e.g. in $\mathbb{R}^{n}$ it is true. Also, when dealing with metric spaces it can sometimes be useful to work with sequential compactness instead of compactness, as they are equivalent. Proving the Heine-Borel theorem also turns out quite straight forward once working with sequences.
Jan
29
answered Complete Metric Spaces
Jan
27
comment Incorrect use of Borel Cantelli?
The theorem seems correct, but I'm having trouble seeing why it seems to you that $X_{n}\rightarrow 0$ pointwise?
Jan
27
comment Incorrect use of Borel Cantelli?
What result are you proving in this theorem? And also, what arguments exactly lead you to think that $X_{n}\rightarrow 0$ a.s.?
Jan
26
answered Integral does not have expected value
Jan
26
comment Does any rotation matrix in 3-d space have only one non-zero eigenvector?
What if the matrix rotates $180^{o}$? Consider 2D and 3D cases.
Jan
26
comment Generator of cyclic groups
So, what group exactly are you dealing with? $\mathbb{Z}_{13}\times\mathbb{Z}_{11}\times\mathbb{Z}_{7}$ ?
Jan
26
comment basis closed, bounded
@Buddy: What do you mean by sequences being closed? That the set $\{x_{n}:n\in\mathbb{N}\}$ is closed, or? And how does that relate to the set itself being closed?
Jan
13
comment On the extension of solutions to ODEs
About the solution of $x'=1+x^{2}$, you probably meant $x(t)=tan(c+t)$?
Jan
12
comment Let $\mathbb{R}^m_+ = \lbrace x \in \mathbb{R}^m : x \geqslant 0 \rbrace.$ What is the boundary set of the set $ \mathbb{R}^m_+$?
The definition of boundary is not $X \backslash X^{0}$. Consider e.g. the rationals $\mathbb{Q}\subset \mathbb{R}$. Its boundary is $\mathbb{R}$, while $\mathbb{Q}^{0}=\emptyset$, since $\mathbb{Q}$ contains no intervals.
Jan
11
accepted Example of a paracompact space that is not metrizable
Jan
11
comment Example of a paracompact space that is not metrizable
Thanks Henno. The Sorgenfrey line seems like a good counter-example for many cases, more than I had suspected.
Jan
11
comment Example of a paracompact space that is not metrizable
Yeah, I did confuse it with $[0,\omega_{1})$. Thanks.
Jan
11
comment Example of a paracompact space that is not metrizable
Thanks, I agree that Compact Hausdorff spaces are paracompact, and I found one good counter example from there. But to be honest, most of the examples in that thread do not convince me. E.g. $[0,\omega_{1}]$ is certainly not a Compact Hausdorff space with the order topology, but sequentially compact which is not equivalent with compactness in general. It is not even Lindelöf. Also, I fail to follow why $\{0,1\}^{\mathbb{R}}$ is not metrizable with the given argument there? But thanks alot, that link was very helpful.
Jan
11
asked Example of a paracompact space that is not metrizable
Jan
11
comment Is this proof about $a^3>a \rightarrow a^5>a $ correct?
@andreas.vitikan: $a^{3}>a^{1}$ does not imply $a\in \mathbb{R}_{+}$. Consider e.g. $a=-\frac{1}{2}$.
Jan
11
awarded  Scholar
Jan
11
accepted Total boundedness, an equivalent expression
Jan
11
accepted Something about boundedness in $\mathbb{R}^{n}$ and also in general..
Jan
11
accepted On the product of separable spaces.