Prescribed set of continuous functions
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Given two sets $X$, $Y$ and a set of functions $F subset Y^X$ such that every constant function from $X$ to $Y$ is in $F$.
Question: Are there topologies on $X$ and $Y$ such that the set $C(X,Y)$ of continuous function is equal to $F$ ?
It is clear that it suffices to find a suitable topology on $Y$ and then take the initial topology on $X$ with respect to all functions in $F$. In general, if we take any topology on $Y$ and initial topology on $X$ with respect to all functions in $F$, we will get $F subset C(X,Y)$. But in many cases we are not able to conclude equality, e.g. if $card(X) = card(Y)$, $F neq Y^X$ but contains some bijective function and Y carries the discrete topology, we will get as initial topology on X again the discrete topology and therefore we have that $C(X,Y) = Y^X neq F$.
A possible approach could be to endow $F$ with some topology and consider $Y$ as a subspace of $F$ (by identifying it with the set of constant functions from $X$ to $Y$). But how to find a convenient topology on $F$ that leads us to $F = C(X,Y)$?
general-topology continuity
asked Dec 3 '18 at 9:59
Daniel W.Daniel W.
162
$endgroup$
add a comment |
$begingroup$
Given two sets $X$, $Y$ and a set of functions $F subset Y^X$ such that every constant function from $X$ to $Y$ is in $F$.
Question: Are there topologies on $X$ and $Y$ such that the set $C(X,Y)$ of continuous function is equal to $F$ ?
It is clear that it suffices to find a suitable topology on $Y$ and then take the initial topology on $X$ with respect to all functions in $F$. In general, if we take any topology on $Y$ and initial topology on $X$ with respect to all functions in $F$, we will get $F subset C(X,Y)$. But in many cases we are not able to conclude equality, e.g. if $card(X) = card(Y)$, $F neq Y^X$ but contains some bijective function and Y carries the discrete topology, we will get as initial topology on X again the discrete topology and therefore we have that $C(X,Y) = Y^X neq F$.
A possible approach could be to endow $F$ with some topology and consider $Y$ as a subspace of $F$ (by identifying it with the set of constant functions from $X$ to $Y$). But how to find a convenient topology on $F$ that leads us to $F = C(X,Y)$?
general-topology continuity
asked Dec 3 '18 at 9:59
Daniel W.Daniel W.
162
$endgroup$
2
$begingroup$
An initial topology is constructed to make e.g. projections continuous in a product but in fact this initial topology then also induces a lot of extra continuous functions, not in a prescribed list. I don't see much hope for this kind of result.
$endgroup$
– Henno Brandsma
Dec 3 '18 at 22:25
add a comment |
$begingroup$
Given two sets $X$, $Y$ and a set of functions $F subset Y^X$ such that every constant function from $X$ to $Y$ is in $F$.
Question: Are there topologies on $X$ and $Y$ such that the set $C(X,Y)$ of continuous function is equal to $F$ ?
It is clear that it suffices to find a suitable topology on $Y$ and then take the initial topology on $X$ with respect to all functions in $F$. In general, if we take any topology on $Y$ and initial topology on $X$ with respect to all functions in $F$, we will get $F subset C(X,Y)$. But in many cases we are not able to conclude equality, e.g. if $card(X) = card(Y)$, $F neq Y^X$ but contains some bijective function and Y carries the discrete topology, we will get as initial topology on X again the discrete topology and therefore we have that $C(X,Y) = Y^X neq F$.
A possible approach could be to endow $F$ with some topology and consider $Y$ as a subspace of $F$ (by identifying it with the set of constant functions from $X$ to $Y$). But how to find a convenient topology on $F$ that leads us to $F = C(X,Y)$?
general-topology continuity
asked Dec 3 '18 at 9:59
Daniel W.Daniel W.
162
$endgroup$
Given two sets $X$, $Y$ and a set of functions $F subset Y^X$ such that every constant function from $X$ to $Y$ is in $F$.
Question: Are there topologies on $X$ and $Y$ such that the set $C(X,Y)$ of continuous function is equal to $F$ ?
It is clear that it suffices to find a suitable topology on $Y$ and then take the initial topology on $X$ with respect to all functions in $F$. In general, if we take any topology on $Y$ and initial topology on $X$ with respect to all functions in $F$, we will get $F subset C(X,Y)$. But in many cases we are not able to conclude equality, e.g. if $card(X) = card(Y)$, $F neq Y^X$ but contains some bijective function and Y carries the discrete topology, we will get as initial topology on X again the discrete topology and therefore we have that $C(X,Y) = Y^X neq F$.
A possible approach could be to endow $F$ with some topology and consider $Y$ as a subspace of $F$ (by identifying it with the set of constant functions from $X$ to $Y$). But how to find a convenient topology on $F$ that leads us to $F = C(X,Y)$?
general-topology continuity
general-topology continuity
asked Dec 3 '18 at 9:59
Daniel W.Daniel W.
162
asked Dec 3 '18 at 9:59
Daniel W.Daniel W.
162
asked Dec 3 '18 at 9:59
Daniel W.Daniel W.
162
asked Dec 3 '18 at 9:59
Daniel W.Daniel W.
162
asked Dec 3 '18 at 9:59
Daniel W.Daniel W.
162
162
2
$begingroup$
An initial topology is constructed to make e.g. projections continuous in a product but in fact this initial topology then also induces a lot of extra continuous functions, not in a prescribed list. I don't see much hope for this kind of result.
$endgroup$
– Henno Brandsma
Dec 3 '18 at 22:25
add a comment |
2
$begingroup$
An initial topology is constructed to make e.g. projections continuous in a product but in fact this initial topology then also induces a lot of extra continuous functions, not in a prescribed list. I don't see much hope for this kind of result.
$endgroup$
– Henno Brandsma
Dec 3 '18 at 22:25
2
2
$begingroup$
An initial topology is constructed to make e.g. projections continuous in a product but in fact this initial topology then also induces a lot of extra continuous functions, not in a prescribed list. I don't see much hope for this kind of result.
$endgroup$
– Henno Brandsma
Dec 3 '18 at 22:25
$begingroup$
An initial topology is constructed to make e.g. projections continuous in a product but in fact this initial topology then also induces a lot of extra continuous functions, not in a prescribed list. I don't see much hope for this kind of result.
$endgroup$
– Henno Brandsma
Dec 3 '18 at 22:25
add a comment |
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$begingroup$
An initial topology is constructed to make e.g. projections continuous in a product but in fact this initial topology then also induces a lot of extra continuous functions, not in a prescribed list. I don't see much hope for this kind of result.
$endgroup$
– Henno Brandsma
Dec 3 '18 at 22:25