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Sep
17
comment $f[A\cup{B}]=f[A]\cup{f[B]}$
If you're worried about how $f$ interacts with the set builder notation, you can avoid it completely by proving $f(A\cup B)\subseteq f(A)\cup f(B)$ and $f(A)\cup f(B)\subseteq f(A\cup B)$. But as has already been said, your existing argument is correct.
Sep
17
answered Explanation on a double summation
Sep
17
reviewed Reopen What is the correct answer to this answered combinatorics problem?
Sep
17
reviewed Leave Open Birational and faithfully flat $\implies$ isomorphism
Sep
17
reviewed Approve Find the tangent to $f(x)$
Sep
17
reviewed No Action Needed $b$ uniformly elliptic and bounded $x\mapsto (b(x))^{-1}$ uniformly elliptic and bounded?
Sep
17
reviewed Close Indexing a vector function, $E(s)=(E_1(s),E_2(s),E_3(s))$, in MATLAB without evaulating the function
Sep
17
answered Find the tangent to $f(x)$
Sep
17
reviewed No Action Needed automorphism of a rooted tree
Sep
16
reviewed Leave Open How to compute $\lim_{n\rightarrow\infty}\frac1n\left\{(2n+1)(2n+2)\cdots(2n+n)\right\}^{1/n}$
Sep
16
reviewed No Action Needed Counting number of solutions for $x = (a-1)(b-2)(c-3)(d-4)(e-5)$
Sep
16
reviewed Approve Maximum of two positive operators
Sep
16
reviewed Looks OK What is the sum of this series?
Sep
14
reviewed Reviewed Prove that the field F is a vector space over itself.
Sep
14
reviewed Close $[E:F]$ can be divided by $|Gal(E/F)| $?
Sep
13
awarded  Good Answer
Sep
10
comment Proof for cyclic permutation
Write $a=(a_1,a_2,\dotsc,a_r)$, let $x$ be an element of $\{1,\dotsc,n\}$, and see what happens when you apply $a$ to $x$ $r$ times.
Sep
10
reviewed Approve Asymptotic bounds. What software to use?
Sep
10
comment Proving that 4 specified sets are not algebraic
For 3 and 4, assume $f$ is a polynomial vanishing on your specified set. Then by restricting to each line through the origin (i.e., set $y=\lambda x$ for some $\lambda$, or $x=0$), you can get a bunch of polynomials of one variable, each of which has infinitely many roots, so must be $0$. Thus you can show that $f$ vanishes everywhere. Possibly a similar method can be adapted to the others.
Sep
10
reviewed Close Twice as much vs Two times as much vs Double