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Doing math.


Mar
13
comment Cantor Function Question
@AsafKaragila: I'm guessing that he might mean the following. Some analysis textbooks define the Cantor function by defining it as constant pieces outside the Cantor set (in the open intervals that were removed at each step), and then inside the Cantor set by limits. In this case, you don't get any expression of the function inside the Cantor set, but only in its complement.
Mar
13
comment Cantor Function Question
Congrats for 1,000 answers.
Mar
13
comment How is Hausdorff Distance sensitive to position?
I only mentioned that condition in order to avoid writing concrete counter-examples. However, the shortest distance also fails to satisfy the triangle-inequality. Take $A=[0,1]$, $B=[2,3]$, and $C=[4,5]$ as a simple example. By shortest distance, $d(A,C)>d(A,B)+d(B,C)$, since $3>2$. I don't think the triangle inequality is negotiable when defining metric spaces. What is there left, the symmetry?
Mar
13
comment How is Hausdorff Distance sensitive to position?
A small note to the second sentence in the last paragraph: the shortest distance is not actually a metric. It fails to satisfy e.g. the condition $d(A,B)=0\Leftrightarrow A=B$.
Mar
13
revised condition $\mu ^*(B\cup C)=\mu ^*(B)+\mu ^*(C)$
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Mar
13
answered condition $\mu ^*(B\cup C)=\mu ^*(B)+\mu ^*(C)$
Mar
13
comment Is there really no way to integrate $e^{-x^2}$?
You probably mean that any continuous function is Riemann integrable on a compact interval.
Mar
12
comment Trying to define $\mathbb{R}^{0.5}$ topologically
Thanks! Interesting way of looking at it.
Mar
12
comment Trying to define $\mathbb{R}^{0.5}$ topologically
Hi Ittay. Maybe this question is just triviality, but could you clarify why $A\times A$ with a single point removed is still connected? I know if $A=\mathbb{R}$ then one can either produce arguments with topological dimensions or use path connectedness. But for any connected space $A$ I can't see how it follows immediately.
Mar
12
revised Open sets in a subspace topology
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Mar
12
revised How is Hausdorff Distance sensitive to position?
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Mar
12
comment How is Hausdorff Distance sensitive to position?
Note that the first definition given for Hausdorff distance is not actually even a distance; the real Hausdorff distance is the "generalized" version. Could you explain more in detail what you mean by the expression: sensitive to position?
Mar
12
comment $\int fg = \lim \int f_n g$ for any $g \in L^q$ and for uniformly bounded $f_n$ in $L^p$
About your second last comment: If you would use DCT to argue the following: $\big| \int f_{n}g-\int fg\big|=\big| \int f_{n}g-fg\big|\leq \|f_{n}g-fg\|_{1}\to 0$, then you would have to show that $|f_{n}g-fg |\leq h$ for some integrable $h$. But do we know that $fg\in L^{1}$?
Mar
12
comment $\int fg = \lim \int f_n g$ for any $g \in L^q$ and for uniformly bounded $f_n$ in $L^p$
It would make sense if $q$ was the Hölder conjugate of $p$. Are you sure there is no typo in the material? Because usually we say that $f_{n}\to f$ weakly in $L^{p}$ if the convergence you wrote applies for all $g\in L^{q}$, where $q$ is the Hölder conjugate of $p$. Also about your current proof, $\|f_{n}-f\|_{p}\to 0$ is not entirely trivial either. You could prove that $f_{n}\to f$ almost everywhere + $\|f_{n}\|\leq M$ for all $n$ implies $f_{n}\to f$ in $L^{p}$ (using dominated convergence theorem). Note that almost everywhere convergence is not sufficient alone to make it happen.
Mar
12
comment $\int fg = \lim \int f_n g$ for any $g \in L^q$ and for uniformly bounded $f_n$ in $L^p$
You probably need to restrict $q$ so that $p^{-1}+q^{-1}=1$, in case you used Hölder in the last inequality.
Mar
12
revised Let $\displaystyle f$ be a continuous real valued function on the metric space $\displaystyle (X,d)$.
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Mar
12
revised Let $\displaystyle f$ be a continuous real valued function on the metric space $\displaystyle (X,d)$.
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Mar
12
answered Let $\displaystyle f$ be a continuous real valued function on the metric space $\displaystyle (X,d)$.
Mar
11
comment Proving $\mu(A)=\inf\{\mu(O) \mid A\subseteq O, O \text{ open}\}$
Are you sure $\mu$ is not a finite measure, or atleast finite on compact sets? If defining for instance $\mu$ on Borel sets of $[0,1]$ s.t. $\mu(\emptyset)=0$, $\mu(\{1\})=1$ and $\mu(A)=+\infty$ for all other Borel sets $A$‚ then $\mu$ is not regular. Reason being that any open set containing $\{1\}$ has infinite measure.
Mar
11
comment Cardinality of the collection of all compact metric spaces
Thanks, again, Brian.