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Jul
22
comment Hausdorff Measure
By line do you mean a subset of $\mathbb{R}^{2}$ or an interval i.e. a subset of $\mathbb{R}$? The answer is different depending on what you mean.
Jul
12
comment Proving a theorem from topology
To me this looks like a definition rather than a theorem. Also, where is the proof that you can not follow?
Jul
4
comment How to solve infinite repeating exponents
@Matt. Dito. And even further, you may notice that $e^{\frac{ln(2)}{2}}=e^{ln(\sqrt{2})}=\sqrt{2}$.
Jul
4
comment Not every metric is induced from a norm
Did you mean instead that any metric not satisfying either of those can not come from a norm? And not the other way around? And also, these two conditions on the metric are sufficient to define a norm..?
Jul
4
comment Contradiction! Any Symbol for?
Where I live, it is standard (and literally everyone uses it) is the symbol $\lightning$. We don't use it on latex typing, but everything that goes on blackboard and personal notes etc.
Jul
3
revised Is this space normal?
edited the topic title and one sentence.
Jul
3
suggested approved edit on Is this space normal?
Jun
29
comment Question about proving something using reverse Fatou's lemma
@madprob: You're right, thanks. And even though I wrote it there, I actually didn't use it in the proof. Indeed $\infty-\infty$ can never occur there as $\mu(g)<\infty$.
Jun
29
revised Question about proving something using reverse Fatou's lemma
took the integrability of $f_{n}$ out.
Jun
28
comment bounded measurable function is the uniform limit of a sequence of simple functions
The key point behind the uniform part of this statement lies in the boundedness of $f$.
Jun
19
comment Difference between supremum and maximum
@copper.hat: The two sentences seem to contradict each other. Correct me if I'm wrong, but first you say maximum is attained and supremum might not, and then you give an example where supremum is attained but maximum isn't. Maybe I misunderstood your point?
Jun
19
comment Subset of $\mathbb{I}\cap [0,1]$ (irrationals in [0,1]) that is closed in $\mathbb{R}$ and has measure $\epsilon \in (0,1)$
In what text did you see this and what page?
Jun
19
comment A simple question about open set
@Mathematics $B(-1,r)=\{x\in S:d(x,-1)<r\}$ as a nhood of $-1$ in $S$ is by definition a subset of $S$. Ofcourse the open ball in $X$ is not a subset of $S$.
Jun
18
comment Continuous images of open sets are Borel?
Thanks Brian, this answer is also very enlightening.
Jun
18
awarded  Nice Question
Jun
18
comment Continuous images of open sets are Borel?
@MichaelGreinecker: Thanks!
Jun
18
comment Continuous images of open sets are Borel?
@t.b. True, that would do the job. Thanks.
Jun
18
accepted Continuous images of open sets are Borel?
Jun
18
asked Continuous images of open sets are Borel?
Jun
17
comment Compactness in $\mathbb{Q}$
@MarcoCastronovo: You can do it the following way. Let $A\subset X$ and give $A$ the subspace topology of $X$. Show that $A$ is compact if and only if every cover of $A$ in $X$ has a finite subcover. The role of $X$ was arbitrary, as long as the topology of $A$ is the subspace topology from $X$.