Subspace topology is the unique topology that satisfies the Characteristic Property
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Lee has as an exercise that the subspace topology is the unique topology that satisfies the characteristic property which is for $Ssubset X$ a function $F:Y to S$ is continuous iff the composition $i_Scirc F$ is continuous.
I think that he means that the only topology that this is true is the subspace topology, but I’m not sure how to prove that no other topology satisfies the property. My first question is what needs to be shown in a proof to say that it is the unique topology, and once I know this I should be able to apply it to the exercise.
general-topology
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
282
$endgroup$
add a comment |
$begingroup$
Lee has as an exercise that the subspace topology is the unique topology that satisfies the characteristic property which is for $Ssubset X$ a function $F:Y to S$ is continuous iff the composition $i_Scirc F$ is continuous.
I think that he means that the only topology that this is true is the subspace topology, but I’m not sure how to prove that no other topology satisfies the property. My first question is what needs to be shown in a proof to say that it is the unique topology, and once I know this I should be able to apply it to the exercise.
general-topology
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
282
$endgroup$
$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03
add a comment |
$begingroup$
Lee has as an exercise that the subspace topology is the unique topology that satisfies the characteristic property which is for $Ssubset X$ a function $F:Y to S$ is continuous iff the composition $i_Scirc F$ is continuous.
I think that he means that the only topology that this is true is the subspace topology, but I’m not sure how to prove that no other topology satisfies the property. My first question is what needs to be shown in a proof to say that it is the unique topology, and once I know this I should be able to apply it to the exercise.
general-topology
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
282
$endgroup$
Lee has as an exercise that the subspace topology is the unique topology that satisfies the characteristic property which is for $Ssubset X$ a function $F:Y to S$ is continuous iff the composition $i_Scirc F$ is continuous.
I think that he means that the only topology that this is true is the subspace topology, but I’m not sure how to prove that no other topology satisfies the property. My first question is what needs to be shown in a proof to say that it is the unique topology, and once I know this I should be able to apply it to the exercise.
general-topology
general-topology
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
282
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
282
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
282
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
282
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
282
282
$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03
add a comment |
$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03
$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03
add a comment |
1 Answer
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Suppose that $tau$ is any topology on $S$ that satisfies the "characteristic property".
Then let $1_S$ be the identity on $(S, tau)$, which is always continuous (for any topology on $S$) so we know that $i_S circ 1_S = i_S$ is continuous (using the "characteristic property" in one direction). This implies that if $U$ is open in $X$ then $i_S^{-1}[U] = U cap S$ is open in $tau$. So the subspace topology $tau_S$ obeys $tau_S subseteq tau$.
Define the function $f(x)=x$ from $(S,tau_S)$ to $(S,tau)$, and note that $i_S = i_S circ f$ as a function $ (S, tau_S) to X$ is continuous (we know the subspace topology $tau_S$ makes this map continuous) and so $f$ is continuous (by the other direction of the "characteristic property"), which implies that for all $U in tau$, $f^{-1}[U]=U in tau_S$ or $tau subseteq tau_S$ and so $tau = tau_S$ and we have unicity of $tau_S$.
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
115k349125
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Suppose that $tau$ is any topology on $S$ that satisfies the "characteristic property".
Then let $1_S$ be the identity on $(S, tau)$, which is always continuous (for any topology on $S$) so we know that $i_S circ 1_S = i_S$ is continuous (using the "characteristic property" in one direction). This implies that if $U$ is open in $X$ then $i_S^{-1}[U] = U cap S$ is open in $tau$. So the subspace topology $tau_S$ obeys $tau_S subseteq tau$.
Define the function $f(x)=x$ from $(S,tau_S)$ to $(S,tau)$, and note that $i_S = i_S circ f$ as a function $ (S, tau_S) to X$ is continuous (we know the subspace topology $tau_S$ makes this map continuous) and so $f$ is continuous (by the other direction of the "characteristic property"), which implies that for all $U in tau$, $f^{-1}[U]=U in tau_S$ or $tau subseteq tau_S$ and so $tau = tau_S$ and we have unicity of $tau_S$.
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
$begingroup$
Suppose that $tau$ is any topology on $S$ that satisfies the "characteristic property".
Then let $1_S$ be the identity on $(S, tau)$, which is always continuous (for any topology on $S$) so we know that $i_S circ 1_S = i_S$ is continuous (using the "characteristic property" in one direction). This implies that if $U$ is open in $X$ then $i_S^{-1}[U] = U cap S$ is open in $tau$. So the subspace topology $tau_S$ obeys $tau_S subseteq tau$.
Define the function $f(x)=x$ from $(S,tau_S)$ to $(S,tau)$, and note that $i_S = i_S circ f$ as a function $ (S, tau_S) to X$ is continuous (we know the subspace topology $tau_S$ makes this map continuous) and so $f$ is continuous (by the other direction of the "characteristic property"), which implies that for all $U in tau$, $f^{-1}[U]=U in tau_S$ or $tau subseteq tau_S$ and so $tau = tau_S$ and we have unicity of $tau_S$.
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
$begingroup$
Suppose that $tau$ is any topology on $S$ that satisfies the "characteristic property".
Then let $1_S$ be the identity on $(S, tau)$, which is always continuous (for any topology on $S$) so we know that $i_S circ 1_S = i_S$ is continuous (using the "characteristic property" in one direction). This implies that if $U$ is open in $X$ then $i_S^{-1}[U] = U cap S$ is open in $tau$. So the subspace topology $tau_S$ obeys $tau_S subseteq tau$.
Define the function $f(x)=x$ from $(S,tau_S)$ to $(S,tau)$, and note that $i_S = i_S circ f$ as a function $ (S, tau_S) to X$ is continuous (we know the subspace topology $tau_S$ makes this map continuous) and so $f$ is continuous (by the other direction of the "characteristic property"), which implies that for all $U in tau$, $f^{-1}[U]=U in tau_S$ or $tau subseteq tau_S$ and so $tau = tau_S$ and we have unicity of $tau_S$.
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
Suppose that $tau$ is any topology on $S$ that satisfies the "characteristic property".
Then let $1_S$ be the identity on $(S, tau)$, which is always continuous (for any topology on $S$) so we know that $i_S circ 1_S = i_S$ is continuous (using the "characteristic property" in one direction). This implies that if $U$ is open in $X$ then $i_S^{-1}[U] = U cap S$ is open in $tau$. So the subspace topology $tau_S$ obeys $tau_S subseteq tau$.
Define the function $f(x)=x$ from $(S,tau_S)$ to $(S,tau)$, and note that $i_S = i_S circ f$ as a function $ (S, tau_S) to X$ is continuous (we know the subspace topology $tau_S$ makes this map continuous) and so $f$ is continuous (by the other direction of the "characteristic property"), which implies that for all $U in tau$, $f^{-1}[U]=U in tau_S$ or $tau subseteq tau_S$ and so $tau = tau_S$ and we have unicity of $tau_S$.
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
115k349125
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
115k349125
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
115k349125
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
115k349125
115k349125
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$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03