Closure of set of polynomials without a constant term term in $R^{R}$
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Let $(mathbb{R}^{mathbb{R}}, p)$ be the space of all functions from $mathbb{R}$ to $mathbb{R}$ with topology of Pointwise convergence. I need to find closure of set of all polynomials without constant term in $mathbb{R}^{mathbb{R}}$.
I dont know how to approach to this problem. Hints?
general-topology
asked Dec 2 '18 at 11:25
chaseperfectionchaseperfection
172
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add a comment |
$begingroup$
Let $(mathbb{R}^{mathbb{R}}, p)$ be the space of all functions from $mathbb{R}$ to $mathbb{R}$ with topology of Pointwise convergence. I need to find closure of set of all polynomials without constant term in $mathbb{R}^{mathbb{R}}$.
I dont know how to approach to this problem. Hints?
general-topology
asked Dec 2 '18 at 11:25
chaseperfectionchaseperfection
172
$endgroup$
add a comment |
$begingroup$
Let $(mathbb{R}^{mathbb{R}}, p)$ be the space of all functions from $mathbb{R}$ to $mathbb{R}$ with topology of Pointwise convergence. I need to find closure of set of all polynomials without constant term in $mathbb{R}^{mathbb{R}}$.
I dont know how to approach to this problem. Hints?
general-topology
asked Dec 2 '18 at 11:25
chaseperfectionchaseperfection
172
$endgroup$
Let $(mathbb{R}^{mathbb{R}}, p)$ be the space of all functions from $mathbb{R}$ to $mathbb{R}$ with topology of Pointwise convergence. I need to find closure of set of all polynomials without constant term in $mathbb{R}^{mathbb{R}}$.
I dont know how to approach to this problem. Hints?
general-topology
general-topology
asked Dec 2 '18 at 11:25
chaseperfectionchaseperfection
172
asked Dec 2 '18 at 11:25
chaseperfectionchaseperfection
172
asked Dec 2 '18 at 11:25
chaseperfectionchaseperfection
172
asked Dec 2 '18 at 11:25
chaseperfectionchaseperfection
172
asked Dec 2 '18 at 11:25
chaseperfectionchaseperfection
172
172
add a comment |
add a comment |
1 Answer
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$begingroup$
You exactly get the set ${f: mathbb{R} to mathbb{R}: f(0) = 0}$, which is pointwise closed as it equals $pi_0^{-1}[{0}]$.
To prove this, think about why the set of all polynomials is dense in your space: finitely many input-output pairs determine a polynomial.
answered Dec 2 '18 at 13:23
Henno BrandsmaHenno Brandsma
110k347116
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You exactly get the set ${f: mathbb{R} to mathbb{R}: f(0) = 0}$, which is pointwise closed as it equals $pi_0^{-1}[{0}]$.
To prove this, think about why the set of all polynomials is dense in your space: finitely many input-output pairs determine a polynomial.
answered Dec 2 '18 at 13:23
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
You exactly get the set ${f: mathbb{R} to mathbb{R}: f(0) = 0}$, which is pointwise closed as it equals $pi_0^{-1}[{0}]$.
To prove this, think about why the set of all polynomials is dense in your space: finitely many input-output pairs determine a polynomial.
answered Dec 2 '18 at 13:23
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
You exactly get the set ${f: mathbb{R} to mathbb{R}: f(0) = 0}$, which is pointwise closed as it equals $pi_0^{-1}[{0}]$.
To prove this, think about why the set of all polynomials is dense in your space: finitely many input-output pairs determine a polynomial.
answered Dec 2 '18 at 13:23
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
You exactly get the set ${f: mathbb{R} to mathbb{R}: f(0) = 0}$, which is pointwise closed as it equals $pi_0^{-1}[{0}]$.
To prove this, think about why the set of all polynomials is dense in your space: finitely many input-output pairs determine a polynomial.
answered Dec 2 '18 at 13:23
Henno BrandsmaHenno Brandsma
110k347116
answered Dec 2 '18 at 13:23
Henno BrandsmaHenno Brandsma
110k347116
answered Dec 2 '18 at 13:23
Henno BrandsmaHenno Brandsma
110k347116
answered Dec 2 '18 at 13:23
Henno BrandsmaHenno Brandsma
110k347116
110k347116
add a comment |
add a comment |
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