5h
awarded Fanatic
1d
comment Compactness of a collection
We have sequence $\{g(p_i)\}_{i=0}^\infty$ converging to $g(p)\in[-\epsilon,\epsilon]\cup[1-\epsilon,1+\epsilon]$ since $[-\epsilon,\epsilon]\cup[1-\epsilon,1+\epsilon]$ is closed. Hence $p\in\mathscr{C}(\epsilon)$. So ${\mathscr{C}(\epsilon)}$ is closed.
1d
comment Compactness of a collection
Made an attempt on closedness. Define $g(p)=\max_{\underline{x}\in\{0,1\}^n}(p(\underline{x})-1_{S_1}(\underline{x}))$ where $1_{S_1}(\underline{x})=1$ if $\underline{x}\in S_1$ else $0$. Consider a sequence of polynomials $\{p_i\}_{i=0}^\infty$ in $\mathscr{C}(\epsilon)$ converging to $p$ in $\overline{\mathscr{C}(\epsilon)}$ (closure of collection $\mathscr{C}(\epsilon)$).
1d
comment Compactness of a collection
Consider coefficients of polynomial in Euclidean space.
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asked Compactness of a collection
1d
revised Is this set possibly compact?
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comment Applications for set theory, ordinal theory, infinite combinatorics and general topology in computer science?
Is it possible to seek a finer topology for polynomial hierarchy alone?
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revised Is this set possibly compact?
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revised Is this set possibly compact?
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revised Is this set possibly compact?
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revised Is this set possibly compact?
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revised Is this set possibly compact?
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revised Is this set possibly compact?
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revised Is this set possibly compact?
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asked Is this set possibly compact?
2d
comment A constrained positive polynomial
Looks like $Q_m(x)=(x-a)^2+1$ should produce a counterexample.
2d
comment A constrained positive polynomial
ok this is almost close. Only thing we need then is an example where we can take $a\geq2$?
2d
comment A constrained positive polynomial
I corrected to $a>1$ and prime is not $1\bmod 4$.
2d
revised A constrained positive polynomial
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2d
comment A constrained positive polynomial
Hmmm yes. that works. Which value of $a$ is $Q(a)$ a prime?
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