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2h
revised The derivative function is not continuous
added 69 characters in body
2h
answered The derivative function is not continuous
17h
comment Finding a point with $f(x)=f'(x)$
There were serious problems with the wording of your question. I took the freedom to edit it, feel free to change it if I misinterpreted what you meant.
17h
revised Finding a point with $f(x)=f'(x)$
added 20 characters in body
17h
revised Finding a point with $f(x)=f'(x)$
Clarification.
18h
accepted How do I know that this is the density of the Chebyshev Points?
18h
accepted Would this be a homology theory?
18h
accepted When does open and connected imply path-connected?
1d
reviewed Close Diagonal lemma logic
1d
accepted Unique extendable functions… Is there a theory?
1d
accepted A manifold such that its boundary is a deformation retract of the manifold itself.
1d
comment Undetermined vs. Undefined
I think there will come a time when we must decide to create a tag of $0/0$ questions or create a bot to delete them, because it is getting quite repetitive.
1d
comment Total derivative of $f(A,B)$ , where $f:M(n,\mathbb{R}) \times M(n,\mathbb{R}) \to M(n,\mathbb{R})$
Evaluate the canonical basis you are considering on the linear maps I provided. If the image of the $i$-th element equal the $i$-th collumn of your matrices, then your answer is correct. I advise you to grasp the concept as I told, though. Many times trying to find the matrix first can be not only troublesome but also not helpful at all for a given problem
1d
answered Total derivative of $f(A,B)$ , where $f:M(n,\mathbb{R}) \times M(n,\mathbb{R}) \to M(n,\mathbb{R})$
2d
comment Can we prove that set of irrational numbers is a set using Zermelo-Fraenkel axioms?
No, $A$ is basically all rational numbers under a given real number. There is a way to define the real numbers with sequences of rational numbers (which are function from $\mathbb{N}$ to $\mathbb{Q}$, hence everything is fine). You take equivalence classes of cauchy sequences essentially.
2d
comment Can we prove that set of irrational numbers is a set using Zermelo-Fraenkel axioms?
Okay, first let's get the real numbers: Take $\mathcal{P}(\mathbb{Q})$. Now, consider the set $R$ of elements $A$ of $\mathcal{P}(\mathbb{Q})$ that satisfy: (i) $A$ is neither empty nor the entire $\mathbb{Q}$ (ii) If $x \in A$, then $A$ contains all elements of $\mathbb{Q}$ that are smaller than $x$ (iii) If $x \in A$, then there exists a rational number greater than $x$ still in $A$. This set $R$ is $\mathbb{R}$. Now, for the irrationals, just take the subset of $\mathbb{R}$ consisting of elements which are not of the form $\{x<r\}$ for some rational $r$.
2d
comment Can we prove that set of irrational numbers is a set using Zermelo-Fraenkel axioms?
The set of reals is a subset of $\mathcal{P}(\mathbb{Q})$, which is a set due to the axioms. Basically, if you have the rationals, the "power set axiom" and the "spefication axiom" gives you the real numbers.
2d
comment Can we prove that set of irrational numbers is a set using Zermelo-Fraenkel axioms?
Naturals = "Axiom" ; Integers = Subset of $\mathbb{N} \times \mathbb{N}$ ; Rationals = Subset of $\mathbb{Z} \times \mathbb{Z}$ ; Reals = Subset of $2^{\mathbb{Q}}$ - See en.wikipedia.org/wiki/Dedekind_cut And, of course, the irrationals are a subset of the real numbers. PS: You got the name wrong.
2d
comment Can we prove that matrix multiplication by its inverse is commutative?
I think he is asking what @pjs36 implies. Even if he isn't, it is a interesting information to be adressed here.
2d
comment $m(E)=0$ or $m(E^c)=0$
An idea: maybe the fact that $m(E)>0 \implies E-E$ contains an open ball centered at zero may be of help.