Mixing
Defining a function as transformation of open sets
In a topological space $(X,tau)$ is it true that a function $f$ is uniquely defined if we define $f$ as transformation of open sets in $X$?
My proof would be like follows (by contradiction)
Suppose $f,g$ are such that for each $mathcal{O} in tau$ we have
$f(mathcal{O}) = g(mathcal{O})$, pick $x in X$, and assume $f(x) neq g(x)$.
Let $mathcal{O}_0 subset mathcal{O}$ an open set containing $x$ and pick $x_1 neq x$ in $mathcal{O}_0$. If we iterate we can construct a sequence of open sets $mathcal{O}_k$ such that
$$
begin{array}{l}
mathcal{O}_k subsetmathcal{O}_{k+1} \
x in mathcal{O}_k ; forall k \
x_k in O_{k-1} - mathcal{O}_k
end{array}
$$
such a sequence converges to $x$, if we assume that $f$ and $g$ are continuous then they will converge to the same point, since the for all $k$ we have $f(mathcal{O}_k) = g(mathcal{O}_k)$.
Is it correct?
general-topology functions
asked Nov 20 '18 at 14:53
user8469759
1,3581616
|
show 9 more comments
In a topological space $(X,tau)$ is it true that a function $f$ is uniquely defined if we define $f$ as transformation of open sets in $X$?
My proof would be like follows (by contradiction)
Suppose $f,g$ are such that for each $mathcal{O} in tau$ we have
$f(mathcal{O}) = g(mathcal{O})$, pick $x in X$, and assume $f(x) neq g(x)$.
Let $mathcal{O}_0 subset mathcal{O}$ an open set containing $x$ and pick $x_1 neq x$ in $mathcal{O}_0$. If we iterate we can construct a sequence of open sets $mathcal{O}_k$ such that
$$
begin{array}{l}
mathcal{O}_k subsetmathcal{O}_{k+1} \
x in mathcal{O}_k ; forall k \
x_k in O_{k-1} - mathcal{O}_k
end{array}
$$
such a sequence converges to $x$, if we assume that $f$ and $g$ are continuous then they will converge to the same point, since the for all $k$ we have $f(mathcal{O}_k) = g(mathcal{O}_k)$.
Is it correct?
general-topology functions
asked Nov 20 '18 at 14:53
user8469759
1,3581616
How do you prove $f(X - left{ x right}) = g(X - left{ x right}) Rightarrow X - f(left{ x right}) = X - g(left{ x right})$?
– Federico
Nov 20 '18 at 14:56
Why is ${x}$ closed?
– Randall
Nov 20 '18 at 14:56
Anyway, the answer is no.
– Randall
Nov 20 '18 at 14:56
1
@Randall let's say that he works in T1 spaces... The implication I pointed out is still wrong. As is the general statement
– Federico
Nov 20 '18 at 14:58
1
Why prove it if it may be false?
– Randall
Nov 20 '18 at 15:09
|
show 9 more comments
In a topological space $(X,tau)$ is it true that a function $f$ is uniquely defined if we define $f$ as transformation of open sets in $X$?
My proof would be like follows (by contradiction)
Suppose $f,g$ are such that for each $mathcal{O} in tau$ we have
$f(mathcal{O}) = g(mathcal{O})$, pick $x in X$, and assume $f(x) neq g(x)$.
Let $mathcal{O}_0 subset mathcal{O}$ an open set containing $x$ and pick $x_1 neq x$ in $mathcal{O}_0$. If we iterate we can construct a sequence of open sets $mathcal{O}_k$ such that
$$
begin{array}{l}
mathcal{O}_k subsetmathcal{O}_{k+1} \
x in mathcal{O}_k ; forall k \
x_k in O_{k-1} - mathcal{O}_k
end{array}
$$
such a sequence converges to $x$, if we assume that $f$ and $g$ are continuous then they will converge to the same point, since the for all $k$ we have $f(mathcal{O}_k) = g(mathcal{O}_k)$.
Is it correct?
general-topology functions
asked Nov 20 '18 at 14:53
user8469759
1,3581616
In a topological space $(X,tau)$ is it true that a function $f$ is uniquely defined if we define $f$ as transformation of open sets in $X$?
My proof would be like follows (by contradiction)
Suppose $f,g$ are such that for each $mathcal{O} in tau$ we have
$f(mathcal{O}) = g(mathcal{O})$, pick $x in X$, and assume $f(x) neq g(x)$.
Let $mathcal{O}_0 subset mathcal{O}$ an open set containing $x$ and pick $x_1 neq x$ in $mathcal{O}_0$. If we iterate we can construct a sequence of open sets $mathcal{O}_k$ such that
$$
begin{array}{l}
mathcal{O}_k subsetmathcal{O}_{k+1} \
x in mathcal{O}_k ; forall k \
x_k in O_{k-1} - mathcal{O}_k
end{array}
$$
such a sequence converges to $x$, if we assume that $f$ and $g$ are continuous then they will converge to the same point, since the for all $k$ we have $f(mathcal{O}_k) = g(mathcal{O}_k)$.
Is it correct?
general-topology functions
general-topology functions
asked Nov 20 '18 at 14:53
user8469759
1,3581616
asked Nov 20 '18 at 14:53
user8469759
1,3581616
edited Nov 20 '18 at 16:24
asked Nov 20 '18 at 14:53
user8469759
1,3581616
asked Nov 20 '18 at 14:53
user8469759
1,3581616
asked Nov 20 '18 at 14:53
user8469759
1,3581616
1,3581616
How do you prove $f(X - left{ x right}) = g(X - left{ x right}) Rightarrow X - f(left{ x right}) = X - g(left{ x right})$?
– Federico
Nov 20 '18 at 14:56
Why is ${x}$ closed?
– Randall
Nov 20 '18 at 14:56
Anyway, the answer is no.
– Randall
Nov 20 '18 at 14:56
1
@Randall let's say that he works in T1 spaces... The implication I pointed out is still wrong. As is the general statement
– Federico
Nov 20 '18 at 14:58
1
Why prove it if it may be false?
– Randall
Nov 20 '18 at 15:09
|
show 9 more comments
How do you prove $f(X - left{ x right}) = g(X - left{ x right}) Rightarrow X - f(left{ x right}) = X - g(left{ x right})$?
– Federico
Nov 20 '18 at 14:56
Why is ${x}$ closed?
– Randall
Nov 20 '18 at 14:56
Anyway, the answer is no.
– Randall
Nov 20 '18 at 14:56
1
@Randall let's say that he works in T1 spaces... The implication I pointed out is still wrong. As is the general statement
– Federico
Nov 20 '18 at 14:58
1
Why prove it if it may be false?
– Randall
Nov 20 '18 at 15:09
How do you prove $f(X - left{ x right}) = g(X - left{ x right}) Rightarrow X - f(left{ x right}) = X - g(left{ x right})$?
– Federico
Nov 20 '18 at 14:56
How do you prove $f(X - left{ x right}) = g(X - left{ x right}) Rightarrow X - f(left{ x right}) = X - g(left{ x right})$?
– Federico
Nov 20 '18 at 14:56
Why is ${x}$ closed?
– Randall
Nov 20 '18 at 14:56
Why is ${x}$ closed?
– Randall
Nov 20 '18 at 14:56
Anyway, the answer is no.
– Randall
Nov 20 '18 at 14:56
Anyway, the answer is no.
– Randall
Nov 20 '18 at 14:56
1
1
@Randall let's say that he works in T1 spaces... The implication I pointed out is still wrong. As is the general statement
– Federico
Nov 20 '18 at 14:58
@Randall let's say that he works in T1 spaces... The implication I pointed out is still wrong. As is the general statement
– Federico
Nov 20 '18 at 14:58
1
1
Why prove it if it may be false?
– Randall
Nov 20 '18 at 15:09
Why prove it if it may be false?
– Randall
Nov 20 '18 at 15:09
|
show 9 more comments
1 Answer
1
active
oldest
votes
Suppose that $Y$ is a Hausdorff space. Suppose that $f,g : X to Y$ are continuous maps for which $f(mathcal{O}) = g(mathcal{O})$ for all open $mathcal{O}$ in $X$. We claim that $f=g$.
Suppose to the contrary that there is a point $a in X$ with $f(a) neq g(a)$. As $Y$ is Hausdorff, we may separate these points: take disjoint open sets $U$ and $V$ in $Y$ containing $f(a)$ and $g(a)$ respectively. From $f(a) in U$ we have $a in f^{-1}(U)$, which is an open set. Hence $g(a) in g(f^{-1}(U))$. But by hypothesis, this says $g(a) in f(f^{-1}(U))$. But $f(f^{-1}(U)) subseteq U$, so we have $g(a) in U$, which contradicts the disjointness of $U$ and $V$.
Edit: A similar argument works in the case that $Y$ is only $T_0$. The definition there is that $Y$ is $T_0$ if for all points $y_1 neq y_2$ in $Y$ there is an open set $U$ containing one point but not the other (and you don't have control over which). If we can enclose $f(a)$ in an open set $U$ not containing $g(a)$ then run the same argument above. You obtain the same contradiction. If not, then we can enclose $g(a)$ in an open set that doesn't contain $f(a)$ and you run the symmetric argument (flip the roles of $f$ and $g$) and still get the same contradiction.
answered Nov 20 '18 at 17:13
Randall
9,12611129
So it is true, we can define functions by how they acts on opensets (with some assumptions).
– user8469759
Nov 20 '18 at 19:59
Out of curiosity, why my sequence argument doesn't work?
– user8469759
Nov 21 '18 at 11:31
@user8469759 It doesn't work because that sequence need not converge.
– Randall
Nov 28 '18 at 1:47
add a comment |
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Suppose that $Y$ is a Hausdorff space. Suppose that $f,g : X to Y$ are continuous maps for which $f(mathcal{O}) = g(mathcal{O})$ for all open $mathcal{O}$ in $X$. We claim that $f=g$.
Suppose to the contrary that there is a point $a in X$ with $f(a) neq g(a)$. As $Y$ is Hausdorff, we may separate these points: take disjoint open sets $U$ and $V$ in $Y$ containing $f(a)$ and $g(a)$ respectively. From $f(a) in U$ we have $a in f^{-1}(U)$, which is an open set. Hence $g(a) in g(f^{-1}(U))$. But by hypothesis, this says $g(a) in f(f^{-1}(U))$. But $f(f^{-1}(U)) subseteq U$, so we have $g(a) in U$, which contradicts the disjointness of $U$ and $V$.
Edit: A similar argument works in the case that $Y$ is only $T_0$. The definition there is that $Y$ is $T_0$ if for all points $y_1 neq y_2$ in $Y$ there is an open set $U$ containing one point but not the other (and you don't have control over which). If we can enclose $f(a)$ in an open set $U$ not containing $g(a)$ then run the same argument above. You obtain the same contradiction. If not, then we can enclose $g(a)$ in an open set that doesn't contain $f(a)$ and you run the symmetric argument (flip the roles of $f$ and $g$) and still get the same contradiction.
answered Nov 20 '18 at 17:13
Randall
9,12611129
So it is true, we can define functions by how they acts on opensets (with some assumptions).
– user8469759
Nov 20 '18 at 19:59
Out of curiosity, why my sequence argument doesn't work?
– user8469759
Nov 21 '18 at 11:31
@user8469759 It doesn't work because that sequence need not converge.
– Randall
Nov 28 '18 at 1:47
add a comment |
Suppose that $Y$ is a Hausdorff space. Suppose that $f,g : X to Y$ are continuous maps for which $f(mathcal{O}) = g(mathcal{O})$ for all open $mathcal{O}$ in $X$. We claim that $f=g$.
Suppose to the contrary that there is a point $a in X$ with $f(a) neq g(a)$. As $Y$ is Hausdorff, we may separate these points: take disjoint open sets $U$ and $V$ in $Y$ containing $f(a)$ and $g(a)$ respectively. From $f(a) in U$ we have $a in f^{-1}(U)$, which is an open set. Hence $g(a) in g(f^{-1}(U))$. But by hypothesis, this says $g(a) in f(f^{-1}(U))$. But $f(f^{-1}(U)) subseteq U$, so we have $g(a) in U$, which contradicts the disjointness of $U$ and $V$.
Edit: A similar argument works in the case that $Y$ is only $T_0$. The definition there is that $Y$ is $T_0$ if for all points $y_1 neq y_2$ in $Y$ there is an open set $U$ containing one point but not the other (and you don't have control over which). If we can enclose $f(a)$ in an open set $U$ not containing $g(a)$ then run the same argument above. You obtain the same contradiction. If not, then we can enclose $g(a)$ in an open set that doesn't contain $f(a)$ and you run the symmetric argument (flip the roles of $f$ and $g$) and still get the same contradiction.
answered Nov 20 '18 at 17:13
Randall
9,12611129
So it is true, we can define functions by how they acts on opensets (with some assumptions).
– user8469759
Nov 20 '18 at 19:59
Out of curiosity, why my sequence argument doesn't work?
– user8469759
Nov 21 '18 at 11:31
@user8469759 It doesn't work because that sequence need not converge.
– Randall
Nov 28 '18 at 1:47
add a comment |
Suppose that $Y$ is a Hausdorff space. Suppose that $f,g : X to Y$ are continuous maps for which $f(mathcal{O}) = g(mathcal{O})$ for all open $mathcal{O}$ in $X$. We claim that $f=g$.
Suppose to the contrary that there is a point $a in X$ with $f(a) neq g(a)$. As $Y$ is Hausdorff, we may separate these points: take disjoint open sets $U$ and $V$ in $Y$ containing $f(a)$ and $g(a)$ respectively. From $f(a) in U$ we have $a in f^{-1}(U)$, which is an open set. Hence $g(a) in g(f^{-1}(U))$. But by hypothesis, this says $g(a) in f(f^{-1}(U))$. But $f(f^{-1}(U)) subseteq U$, so we have $g(a) in U$, which contradicts the disjointness of $U$ and $V$.
Edit: A similar argument works in the case that $Y$ is only $T_0$. The definition there is that $Y$ is $T_0$ if for all points $y_1 neq y_2$ in $Y$ there is an open set $U$ containing one point but not the other (and you don't have control over which). If we can enclose $f(a)$ in an open set $U$ not containing $g(a)$ then run the same argument above. You obtain the same contradiction. If not, then we can enclose $g(a)$ in an open set that doesn't contain $f(a)$ and you run the symmetric argument (flip the roles of $f$ and $g$) and still get the same contradiction.
answered Nov 20 '18 at 17:13
Randall
9,12611129
Suppose that $Y$ is a Hausdorff space. Suppose that $f,g : X to Y$ are continuous maps for which $f(mathcal{O}) = g(mathcal{O})$ for all open $mathcal{O}$ in $X$. We claim that $f=g$.
Suppose to the contrary that there is a point $a in X$ with $f(a) neq g(a)$. As $Y$ is Hausdorff, we may separate these points: take disjoint open sets $U$ and $V$ in $Y$ containing $f(a)$ and $g(a)$ respectively. From $f(a) in U$ we have $a in f^{-1}(U)$, which is an open set. Hence $g(a) in g(f^{-1}(U))$. But by hypothesis, this says $g(a) in f(f^{-1}(U))$. But $f(f^{-1}(U)) subseteq U$, so we have $g(a) in U$, which contradicts the disjointness of $U$ and $V$.
Edit: A similar argument works in the case that $Y$ is only $T_0$. The definition there is that $Y$ is $T_0$ if for all points $y_1 neq y_2$ in $Y$ there is an open set $U$ containing one point but not the other (and you don't have control over which). If we can enclose $f(a)$ in an open set $U$ not containing $g(a)$ then run the same argument above. You obtain the same contradiction. If not, then we can enclose $g(a)$ in an open set that doesn't contain $f(a)$ and you run the symmetric argument (flip the roles of $f$ and $g$) and still get the same contradiction.
answered Nov 20 '18 at 17:13
Randall
9,12611129
edited Nov 20 '18 at 17:32
answered Nov 20 '18 at 17:13
Randall
9,12611129
answered Nov 20 '18 at 17:13
Randall
9,12611129
answered Nov 20 '18 at 17:13
Randall
9,12611129
9,12611129
So it is true, we can define functions by how they acts on opensets (with some assumptions).
– user8469759
Nov 20 '18 at 19:59
Out of curiosity, why my sequence argument doesn't work?
– user8469759
Nov 21 '18 at 11:31
@user8469759 It doesn't work because that sequence need not converge.
– Randall
Nov 28 '18 at 1:47
add a comment |
So it is true, we can define functions by how they acts on opensets (with some assumptions).
– user8469759
Nov 20 '18 at 19:59
Out of curiosity, why my sequence argument doesn't work?
– user8469759
Nov 21 '18 at 11:31
@user8469759 It doesn't work because that sequence need not converge.
– Randall
Nov 28 '18 at 1:47
So it is true, we can define functions by how they acts on opensets (with some assumptions).
– user8469759
Nov 20 '18 at 19:59
So it is true, we can define functions by how they acts on opensets (with some assumptions).
– user8469759
Nov 20 '18 at 19:59
Out of curiosity, why my sequence argument doesn't work?
– user8469759
Nov 21 '18 at 11:31
Out of curiosity, why my sequence argument doesn't work?
– user8469759
Nov 21 '18 at 11:31
@user8469759 It doesn't work because that sequence need not converge.
– Randall
Nov 28 '18 at 1:47
@user8469759 It doesn't work because that sequence need not converge.
– Randall
Nov 28 '18 at 1:47
add a comment |
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How do you prove $f(X - left{ x right}) = g(X - left{ x right}) Rightarrow X - f(left{ x right}) = X - g(left{ x right})$?
– Federico
Nov 20 '18 at 14:56
Why is ${x}$ closed?
– Randall
Nov 20 '18 at 14:56
Anyway, the answer is no.
– Randall
Nov 20 '18 at 14:56
1
@Randall let's say that he works in T1 spaces... The implication I pointed out is still wrong. As is the general statement
– Federico
Nov 20 '18 at 14:58
1
Why prove it if it may be false?
– Randall
Nov 20 '18 at 15:09