Showing that two mappings are continuous. Introduction to topology
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I have a course were we are learning about topology and have to show in two different exercises that two different mappings are continuous.
I think I have a solution. However, as these concepts a new to me, I would like to ask if this is indeed correct and if there is a better way to solve them.
The first problem:
Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and define $d_{Xtimes Y}:(Xtimes Y)times (Xtimes Y) rightarrow mathbb{R}^+_0$ by
$d_{Xtimes Y}((x_1,y_1),(x_2,y_2))=max(d_X(x_1,x_2),d_Y(y_1,y_2))$
Then show the projection
$p_X:Xtimes Y rightarrow X, p_X(x,y)=x$
is continuous.
I would then solve this by showing that for every open set V in X then $p_X^{-1}(V)$ is an open set in $Xtimes Y$.
So in mathematical terms
$V subseteq X$ and
$forall xin V, exists delta_V >0 : B^X_{delta_V}(x)subseteq V$
and $B^X_{delta_V}(x)={yin X|d_X(x,y)<delta_V}$
We then need there to exist a $delta_{Xtimes Y}$ so that $forall (x,y)in p_X^{-1}(V), exists delta_{Xtimes Y} >0 : B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))subseteq p_X^{-1}(V)$
Where $p_X^{-1}(V)=Vtimes Y$ as the second coordinate in $(x,y)$ is just dropped and can therefore be anything.
As every $x_1 in V$ has a value $delta_X$ we can now look at every $(x_1,y)in p_X^{-1}(V)$, where it is the same element $x_1$ as before, then $B^{Xtimes Y}_{delta_{Xtimes Y}}((x_1,y))subseteq p_X^{-1}(V)$ if $0<delta_{Xtimes Y} leq delta_X$.
I conclude this as for $B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))nsubseteq p_X^{-1}(V)$ then there has to be a point $(x_2,y_2)$ where $d_X(x_1,x_2)<delta_{Xtimes Y}leqdelta_X$ and $x_2notin V$ as every possible $y_2$ is in $p_X^{-1}(V)=Vtimes Y$. But this is contradictory to what we know as $delta_X$ is chosen so every $x_2$ where $d_X(x1,x2)<delta_X$ is in V.
We can therefore conclude that for every open subset of X, called $V$, then $p_X^{-1}(V)$ is an open subset of $X times Y$ as $forall (x,y)in p_X^{-1}(V), exists delta_{Xtimes Y} >0 : B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))subseteq p_X^{-1}(V)$ is fulfilled and the projection is therefore continuous.
The second problem is so similar that it is not currently added.
real-analysis general-topology continuity
$endgroup$
add a comment |
$begingroup$
I have a course were we are learning about topology and have to show in two different exercises that two different mappings are continuous.
I think I have a solution. However, as these concepts a new to me, I would like to ask if this is indeed correct and if there is a better way to solve them.
The first problem:
Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and define $d_{Xtimes Y}:(Xtimes Y)times (Xtimes Y) rightarrow mathbb{R}^+_0$ by
$d_{Xtimes Y}((x_1,y_1),(x_2,y_2))=max(d_X(x_1,x_2),d_Y(y_1,y_2))$
Then show the projection
$p_X:Xtimes Y rightarrow X, p_X(x,y)=x$
is continuous.
I would then solve this by showing that for every open set V in X then $p_X^{-1}(V)$ is an open set in $Xtimes Y$.
So in mathematical terms
$V subseteq X$ and
$forall xin V, exists delta_V >0 : B^X_{delta_V}(x)subseteq V$
and $B^X_{delta_V}(x)={yin X|d_X(x,y)<delta_V}$
We then need there to exist a $delta_{Xtimes Y}$ so that $forall (x,y)in p_X^{-1}(V), exists delta_{Xtimes Y} >0 : B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))subseteq p_X^{-1}(V)$
Where $p_X^{-1}(V)=Vtimes Y$ as the second coordinate in $(x,y)$ is just dropped and can therefore be anything.
As every $x_1 in V$ has a value $delta_X$ we can now look at every $(x_1,y)in p_X^{-1}(V)$, where it is the same element $x_1$ as before, then $B^{Xtimes Y}_{delta_{Xtimes Y}}((x_1,y))subseteq p_X^{-1}(V)$ if $0<delta_{Xtimes Y} leq delta_X$.
I conclude this as for $B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))nsubseteq p_X^{-1}(V)$ then there has to be a point $(x_2,y_2)$ where $d_X(x_1,x_2)<delta_{Xtimes Y}leqdelta_X$ and $x_2notin V$ as every possible $y_2$ is in $p_X^{-1}(V)=Vtimes Y$. But this is contradictory to what we know as $delta_X$ is chosen so every $x_2$ where $d_X(x1,x2)<delta_X$ is in V.
We can therefore conclude that for every open subset of X, called $V$, then $p_X^{-1}(V)$ is an open subset of $X times Y$ as $forall (x,y)in p_X^{-1}(V), exists delta_{Xtimes Y} >0 : B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))subseteq p_X^{-1}(V)$ is fulfilled and the projection is therefore continuous.
The second problem is so similar that it is not currently added.
real-analysis general-topology continuity
$endgroup$
add a comment |
$begingroup$
I have a course were we are learning about topology and have to show in two different exercises that two different mappings are continuous.
I think I have a solution. However, as these concepts a new to me, I would like to ask if this is indeed correct and if there is a better way to solve them.
The first problem:
Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and define $d_{Xtimes Y}:(Xtimes Y)times (Xtimes Y) rightarrow mathbb{R}^+_0$ by
$d_{Xtimes Y}((x_1,y_1),(x_2,y_2))=max(d_X(x_1,x_2),d_Y(y_1,y_2))$
Then show the projection
$p_X:Xtimes Y rightarrow X, p_X(x,y)=x$
is continuous.
I would then solve this by showing that for every open set V in X then $p_X^{-1}(V)$ is an open set in $Xtimes Y$.
So in mathematical terms
$V subseteq X$ and
$forall xin V, exists delta_V >0 : B^X_{delta_V}(x)subseteq V$
and $B^X_{delta_V}(x)={yin X|d_X(x,y)<delta_V}$
We then need there to exist a $delta_{Xtimes Y}$ so that $forall (x,y)in p_X^{-1}(V), exists delta_{Xtimes Y} >0 : B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))subseteq p_X^{-1}(V)$
Where $p_X^{-1}(V)=Vtimes Y$ as the second coordinate in $(x,y)$ is just dropped and can therefore be anything.
As every $x_1 in V$ has a value $delta_X$ we can now look at every $(x_1,y)in p_X^{-1}(V)$, where it is the same element $x_1$ as before, then $B^{Xtimes Y}_{delta_{Xtimes Y}}((x_1,y))subseteq p_X^{-1}(V)$ if $0<delta_{Xtimes Y} leq delta_X$.
I conclude this as for $B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))nsubseteq p_X^{-1}(V)$ then there has to be a point $(x_2,y_2)$ where $d_X(x_1,x_2)<delta_{Xtimes Y}leqdelta_X$ and $x_2notin V$ as every possible $y_2$ is in $p_X^{-1}(V)=Vtimes Y$. But this is contradictory to what we know as $delta_X$ is chosen so every $x_2$ where $d_X(x1,x2)<delta_X$ is in V.
We can therefore conclude that for every open subset of X, called $V$, then $p_X^{-1}(V)$ is an open subset of $X times Y$ as $forall (x,y)in p_X^{-1}(V), exists delta_{Xtimes Y} >0 : B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))subseteq p_X^{-1}(V)$ is fulfilled and the projection is therefore continuous.
The second problem is so similar that it is not currently added.
real-analysis general-topology continuity
$endgroup$
I have a course were we are learning about topology and have to show in two different exercises that two different mappings are continuous.
I think I have a solution. However, as these concepts a new to me, I would like to ask if this is indeed correct and if there is a better way to solve them.
The first problem:
Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and define $d_{Xtimes Y}:(Xtimes Y)times (Xtimes Y) rightarrow mathbb{R}^+_0$ by
$d_{Xtimes Y}((x_1,y_1),(x_2,y_2))=max(d_X(x_1,x_2),d_Y(y_1,y_2))$
Then show the projection
$p_X:Xtimes Y rightarrow X, p_X(x,y)=x$
is continuous.
I would then solve this by showing that for every open set V in X then $p_X^{-1}(V)$ is an open set in $Xtimes Y$.
So in mathematical terms
$V subseteq X$ and
$forall xin V, exists delta_V >0 : B^X_{delta_V}(x)subseteq V$
and $B^X_{delta_V}(x)={yin X|d_X(x,y)<delta_V}$
We then need there to exist a $delta_{Xtimes Y}$ so that $forall (x,y)in p_X^{-1}(V), exists delta_{Xtimes Y} >0 : B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))subseteq p_X^{-1}(V)$
Where $p_X^{-1}(V)=Vtimes Y$ as the second coordinate in $(x,y)$ is just dropped and can therefore be anything.
As every $x_1 in V$ has a value $delta_X$ we can now look at every $(x_1,y)in p_X^{-1}(V)$, where it is the same element $x_1$ as before, then $B^{Xtimes Y}_{delta_{Xtimes Y}}((x_1,y))subseteq p_X^{-1}(V)$ if $0<delta_{Xtimes Y} leq delta_X$.
I conclude this as for $B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))nsubseteq p_X^{-1}(V)$ then there has to be a point $(x_2,y_2)$ where $d_X(x_1,x_2)<delta_{Xtimes Y}leqdelta_X$ and $x_2notin V$ as every possible $y_2$ is in $p_X^{-1}(V)=Vtimes Y$. But this is contradictory to what we know as $delta_X$ is chosen so every $x_2$ where $d_X(x1,x2)<delta_X$ is in V.
We can therefore conclude that for every open subset of X, called $V$, then $p_X^{-1}(V)$ is an open subset of $X times Y$ as $forall (x,y)in p_X^{-1}(V), exists delta_{Xtimes Y} >0 : B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))subseteq p_X^{-1}(V)$ is fulfilled and the projection is therefore continuous.
The second problem is so similar that it is not currently added.
real-analysis general-topology continuity
real-analysis general-topology continuity
asked Jan 9 at 7:45
Andreas PAndreas P
223
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2 Answers
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It suffices to show that for any $(x,y)in p_X^{-1}V$ there exists an open ball $A=B_{d_{Xtimes Y}}((x,y),r)$ centered at $(x,y),$ with $r>0,$ such that $Asubseteq p_X^{-1}V.$
For any $x'in X$ we have $p_X^{-1}{x'}={x'}times Y.$ So for any $Usubset X $ we have $p_X^{-1}U=cup_{x'in U}({x'}times Y)=Utimes Y.$
So if $(x,y)in p_X^{-1}V=Vtimes Y$ then $xin V,$ and since $V$ is open in $X,$ there exists $r>0$ such that $B_{d_X}(x,r)subset V.$ Therefore $$ B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)subseteq$$ $$subseteq B_{d_X}(x,r)times Ysubseteq$$ $$subseteq Vtimes Y=$$ $$=p_X^{-1}V.$$
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There are many "different-looking" metrics on $Xtimes Y$ that generate the same topology that $d_{Xtimes Y}$ does, including $e((x,y),(x',y'))=d_X(x,x')+d_Y(y,y')$ and $f((x,y),(x',y'))=sqrt {d_X(x,x')^2+d_Y(y,y')^2};.$
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– DanielWainfleet
Jan 9 at 10:56
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Would what I did be wrong? Yours is more elegant, but in some ways it seems to be the "same" that I am trying to say as what you are saying. And just for my understanding. $B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)$, you can make the equality because $d_{Xtimes Y}$ uses maximum and if it was one of your examples $e$ or $f$ then it would be '$subseteq$' and not '$=$'?
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– Andreas P
Jan 9 at 11:20
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I dk whether you made a mistake. I haven't examined your details. I just wanted to show a simple proof, for comparison. Regarding $e$ and $f, $ you are right. And (obviously) we can interchange $X$ with $Y.$ So the (Tychonoff) product topology on $Xtimes Y,$ which is the weakest ($subset$-smallest) topology on $Xtimes Y$ such that $ p_X $ and $ p_Y $ are continuous, is weaker (i.e. a subset of) the topologies generated by $d,e, $ or $f$ . In fact all $4$ of them are the same topology.
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– DanielWainfleet
Jan 9 at 23:10
add a comment |
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The projection p:X×Y -> X, (x,y) -> x is continuous.
Proof. Let U be an open subset of X.
Then p$^{-1}$(U) = U×Y is open subset of X×Y
since it is the product of two open sets.
That is true of all topological spaces.
For your problem simply show that the function
d = d$_{X×Y}$ is a metric and U×Y is open when U is.
Actually, the fruitful proposition is to prove is that d
generates the product topology - whenever U is open within
X and V is open within Y, then U×V is open within X×Y.
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I have shown that d is a metric. Would my original post amount to showing $U times Y$ is open given a bit of rewording? Otherwise, I cannot see how this would be done.
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– Andreas P
Jan 9 at 10:40
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The topology on $Xtimes Y$ in the Q is defined by the metric $d_{Xtimes Y}.$ Although it is true that this co-incides with the Tychonoff product topology, this is not stated in the Q, which seems to be the first part of an exercise to show that these topologies are equal. Only after that can we say " $p^{-1}U$ is open in the metric topology because it is the product of 2 open sets".
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– DanielWainfleet
Jan 9 at 10:46
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It suffices to show that for any $(x,y)in p_X^{-1}V$ there exists an open ball $A=B_{d_{Xtimes Y}}((x,y),r)$ centered at $(x,y),$ with $r>0,$ such that $Asubseteq p_X^{-1}V.$
For any $x'in X$ we have $p_X^{-1}{x'}={x'}times Y.$ So for any $Usubset X $ we have $p_X^{-1}U=cup_{x'in U}({x'}times Y)=Utimes Y.$
So if $(x,y)in p_X^{-1}V=Vtimes Y$ then $xin V,$ and since $V$ is open in $X,$ there exists $r>0$ such that $B_{d_X}(x,r)subset V.$ Therefore $$ B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)subseteq$$ $$subseteq B_{d_X}(x,r)times Ysubseteq$$ $$subseteq Vtimes Y=$$ $$=p_X^{-1}V.$$
$endgroup$
$begingroup$
There are many "different-looking" metrics on $Xtimes Y$ that generate the same topology that $d_{Xtimes Y}$ does, including $e((x,y),(x',y'))=d_X(x,x')+d_Y(y,y')$ and $f((x,y),(x',y'))=sqrt {d_X(x,x')^2+d_Y(y,y')^2};.$
$endgroup$
– DanielWainfleet
Jan 9 at 10:56
$begingroup$
Would what I did be wrong? Yours is more elegant, but in some ways it seems to be the "same" that I am trying to say as what you are saying. And just for my understanding. $B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)$, you can make the equality because $d_{Xtimes Y}$ uses maximum and if it was one of your examples $e$ or $f$ then it would be '$subseteq$' and not '$=$'?
$endgroup$
– Andreas P
Jan 9 at 11:20
$begingroup$
I dk whether you made a mistake. I haven't examined your details. I just wanted to show a simple proof, for comparison. Regarding $e$ and $f, $ you are right. And (obviously) we can interchange $X$ with $Y.$ So the (Tychonoff) product topology on $Xtimes Y,$ which is the weakest ($subset$-smallest) topology on $Xtimes Y$ such that $ p_X $ and $ p_Y $ are continuous, is weaker (i.e. a subset of) the topologies generated by $d,e, $ or $f$ . In fact all $4$ of them are the same topology.
$endgroup$
– DanielWainfleet
Jan 9 at 23:10
add a comment |
$begingroup$
It suffices to show that for any $(x,y)in p_X^{-1}V$ there exists an open ball $A=B_{d_{Xtimes Y}}((x,y),r)$ centered at $(x,y),$ with $r>0,$ such that $Asubseteq p_X^{-1}V.$
For any $x'in X$ we have $p_X^{-1}{x'}={x'}times Y.$ So for any $Usubset X $ we have $p_X^{-1}U=cup_{x'in U}({x'}times Y)=Utimes Y.$
So if $(x,y)in p_X^{-1}V=Vtimes Y$ then $xin V,$ and since $V$ is open in $X,$ there exists $r>0$ such that $B_{d_X}(x,r)subset V.$ Therefore $$ B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)subseteq$$ $$subseteq B_{d_X}(x,r)times Ysubseteq$$ $$subseteq Vtimes Y=$$ $$=p_X^{-1}V.$$
$endgroup$
$begingroup$
There are many "different-looking" metrics on $Xtimes Y$ that generate the same topology that $d_{Xtimes Y}$ does, including $e((x,y),(x',y'))=d_X(x,x')+d_Y(y,y')$ and $f((x,y),(x',y'))=sqrt {d_X(x,x')^2+d_Y(y,y')^2};.$
$endgroup$
– DanielWainfleet
Jan 9 at 10:56
$begingroup$
Would what I did be wrong? Yours is more elegant, but in some ways it seems to be the "same" that I am trying to say as what you are saying. And just for my understanding. $B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)$, you can make the equality because $d_{Xtimes Y}$ uses maximum and if it was one of your examples $e$ or $f$ then it would be '$subseteq$' and not '$=$'?
$endgroup$
– Andreas P
Jan 9 at 11:20
$begingroup$
I dk whether you made a mistake. I haven't examined your details. I just wanted to show a simple proof, for comparison. Regarding $e$ and $f, $ you are right. And (obviously) we can interchange $X$ with $Y.$ So the (Tychonoff) product topology on $Xtimes Y,$ which is the weakest ($subset$-smallest) topology on $Xtimes Y$ such that $ p_X $ and $ p_Y $ are continuous, is weaker (i.e. a subset of) the topologies generated by $d,e, $ or $f$ . In fact all $4$ of them are the same topology.
$endgroup$
– DanielWainfleet
Jan 9 at 23:10
add a comment |
$begingroup$
It suffices to show that for any $(x,y)in p_X^{-1}V$ there exists an open ball $A=B_{d_{Xtimes Y}}((x,y),r)$ centered at $(x,y),$ with $r>0,$ such that $Asubseteq p_X^{-1}V.$
For any $x'in X$ we have $p_X^{-1}{x'}={x'}times Y.$ So for any $Usubset X $ we have $p_X^{-1}U=cup_{x'in U}({x'}times Y)=Utimes Y.$
So if $(x,y)in p_X^{-1}V=Vtimes Y$ then $xin V,$ and since $V$ is open in $X,$ there exists $r>0$ such that $B_{d_X}(x,r)subset V.$ Therefore $$ B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)subseteq$$ $$subseteq B_{d_X}(x,r)times Ysubseteq$$ $$subseteq Vtimes Y=$$ $$=p_X^{-1}V.$$
$endgroup$
It suffices to show that for any $(x,y)in p_X^{-1}V$ there exists an open ball $A=B_{d_{Xtimes Y}}((x,y),r)$ centered at $(x,y),$ with $r>0,$ such that $Asubseteq p_X^{-1}V.$
For any $x'in X$ we have $p_X^{-1}{x'}={x'}times Y.$ So for any $Usubset X $ we have $p_X^{-1}U=cup_{x'in U}({x'}times Y)=Utimes Y.$
So if $(x,y)in p_X^{-1}V=Vtimes Y$ then $xin V,$ and since $V$ is open in $X,$ there exists $r>0$ such that $B_{d_X}(x,r)subset V.$ Therefore $$ B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)subseteq$$ $$subseteq B_{d_X}(x,r)times Ysubseteq$$ $$subseteq Vtimes Y=$$ $$=p_X^{-1}V.$$
answered Jan 9 at 10:35
DanielWainfleetDanielWainfleet
35.9k31648
35.9k31648
$begingroup$
There are many "different-looking" metrics on $Xtimes Y$ that generate the same topology that $d_{Xtimes Y}$ does, including $e((x,y),(x',y'))=d_X(x,x')+d_Y(y,y')$ and $f((x,y),(x',y'))=sqrt {d_X(x,x')^2+d_Y(y,y')^2};.$
$endgroup$
– DanielWainfleet
Jan 9 at 10:56
$begingroup$
Would what I did be wrong? Yours is more elegant, but in some ways it seems to be the "same" that I am trying to say as what you are saying. And just for my understanding. $B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)$, you can make the equality because $d_{Xtimes Y}$ uses maximum and if it was one of your examples $e$ or $f$ then it would be '$subseteq$' and not '$=$'?
$endgroup$
– Andreas P
Jan 9 at 11:20
$begingroup$
I dk whether you made a mistake. I haven't examined your details. I just wanted to show a simple proof, for comparison. Regarding $e$ and $f, $ you are right. And (obviously) we can interchange $X$ with $Y.$ So the (Tychonoff) product topology on $Xtimes Y,$ which is the weakest ($subset$-smallest) topology on $Xtimes Y$ such that $ p_X $ and $ p_Y $ are continuous, is weaker (i.e. a subset of) the topologies generated by $d,e, $ or $f$ . In fact all $4$ of them are the same topology.
$endgroup$
– DanielWainfleet
Jan 9 at 23:10
add a comment |
$begingroup$
There are many "different-looking" metrics on $Xtimes Y$ that generate the same topology that $d_{Xtimes Y}$ does, including $e((x,y),(x',y'))=d_X(x,x')+d_Y(y,y')$ and $f((x,y),(x',y'))=sqrt {d_X(x,x')^2+d_Y(y,y')^2};.$
$endgroup$
– DanielWainfleet
Jan 9 at 10:56
$begingroup$
Would what I did be wrong? Yours is more elegant, but in some ways it seems to be the "same" that I am trying to say as what you are saying. And just for my understanding. $B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)$, you can make the equality because $d_{Xtimes Y}$ uses maximum and if it was one of your examples $e$ or $f$ then it would be '$subseteq$' and not '$=$'?
$endgroup$
– Andreas P
Jan 9 at 11:20
$begingroup$
I dk whether you made a mistake. I haven't examined your details. I just wanted to show a simple proof, for comparison. Regarding $e$ and $f, $ you are right. And (obviously) we can interchange $X$ with $Y.$ So the (Tychonoff) product topology on $Xtimes Y,$ which is the weakest ($subset$-smallest) topology on $Xtimes Y$ such that $ p_X $ and $ p_Y $ are continuous, is weaker (i.e. a subset of) the topologies generated by $d,e, $ or $f$ . In fact all $4$ of them are the same topology.
$endgroup$
– DanielWainfleet
Jan 9 at 23:10
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There are many "different-looking" metrics on $Xtimes Y$ that generate the same topology that $d_{Xtimes Y}$ does, including $e((x,y),(x',y'))=d_X(x,x')+d_Y(y,y')$ and $f((x,y),(x',y'))=sqrt {d_X(x,x')^2+d_Y(y,y')^2};.$
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– DanielWainfleet
Jan 9 at 10:56
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There are many "different-looking" metrics on $Xtimes Y$ that generate the same topology that $d_{Xtimes Y}$ does, including $e((x,y),(x',y'))=d_X(x,x')+d_Y(y,y')$ and $f((x,y),(x',y'))=sqrt {d_X(x,x')^2+d_Y(y,y')^2};.$
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– DanielWainfleet
Jan 9 at 10:56
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Would what I did be wrong? Yours is more elegant, but in some ways it seems to be the "same" that I am trying to say as what you are saying. And just for my understanding. $B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)$, you can make the equality because $d_{Xtimes Y}$ uses maximum and if it was one of your examples $e$ or $f$ then it would be '$subseteq$' and not '$=$'?
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– Andreas P
Jan 9 at 11:20
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Would what I did be wrong? Yours is more elegant, but in some ways it seems to be the "same" that I am trying to say as what you are saying. And just for my understanding. $B_{d_{Xtimes Y}}((x,y),r)=B_{d_X}(x,r)times B_{d_Y}(y,r)$, you can make the equality because $d_{Xtimes Y}$ uses maximum and if it was one of your examples $e$ or $f$ then it would be '$subseteq$' and not '$=$'?
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– Andreas P
Jan 9 at 11:20
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I dk whether you made a mistake. I haven't examined your details. I just wanted to show a simple proof, for comparison. Regarding $e$ and $f, $ you are right. And (obviously) we can interchange $X$ with $Y.$ So the (Tychonoff) product topology on $Xtimes Y,$ which is the weakest ($subset$-smallest) topology on $Xtimes Y$ such that $ p_X $ and $ p_Y $ are continuous, is weaker (i.e. a subset of) the topologies generated by $d,e, $ or $f$ . In fact all $4$ of them are the same topology.
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– DanielWainfleet
Jan 9 at 23:10
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I dk whether you made a mistake. I haven't examined your details. I just wanted to show a simple proof, for comparison. Regarding $e$ and $f, $ you are right. And (obviously) we can interchange $X$ with $Y.$ So the (Tychonoff) product topology on $Xtimes Y,$ which is the weakest ($subset$-smallest) topology on $Xtimes Y$ such that $ p_X $ and $ p_Y $ are continuous, is weaker (i.e. a subset of) the topologies generated by $d,e, $ or $f$ . In fact all $4$ of them are the same topology.
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– DanielWainfleet
Jan 9 at 23:10
add a comment |
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The projection p:X×Y -> X, (x,y) -> x is continuous.
Proof. Let U be an open subset of X.
Then p$^{-1}$(U) = U×Y is open subset of X×Y
since it is the product of two open sets.
That is true of all topological spaces.
For your problem simply show that the function
d = d$_{X×Y}$ is a metric and U×Y is open when U is.
Actually, the fruitful proposition is to prove is that d
generates the product topology - whenever U is open within
X and V is open within Y, then U×V is open within X×Y.
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I have shown that d is a metric. Would my original post amount to showing $U times Y$ is open given a bit of rewording? Otherwise, I cannot see how this would be done.
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– Andreas P
Jan 9 at 10:40
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The topology on $Xtimes Y$ in the Q is defined by the metric $d_{Xtimes Y}.$ Although it is true that this co-incides with the Tychonoff product topology, this is not stated in the Q, which seems to be the first part of an exercise to show that these topologies are equal. Only after that can we say " $p^{-1}U$ is open in the metric topology because it is the product of 2 open sets".
$endgroup$
– DanielWainfleet
Jan 9 at 10:46
add a comment |
$begingroup$
The projection p:X×Y -> X, (x,y) -> x is continuous.
Proof. Let U be an open subset of X.
Then p$^{-1}$(U) = U×Y is open subset of X×Y
since it is the product of two open sets.
That is true of all topological spaces.
For your problem simply show that the function
d = d$_{X×Y}$ is a metric and U×Y is open when U is.
Actually, the fruitful proposition is to prove is that d
generates the product topology - whenever U is open within
X and V is open within Y, then U×V is open within X×Y.
$endgroup$
$begingroup$
I have shown that d is a metric. Would my original post amount to showing $U times Y$ is open given a bit of rewording? Otherwise, I cannot see how this would be done.
$endgroup$
– Andreas P
Jan 9 at 10:40
$begingroup$
The topology on $Xtimes Y$ in the Q is defined by the metric $d_{Xtimes Y}.$ Although it is true that this co-incides with the Tychonoff product topology, this is not stated in the Q, which seems to be the first part of an exercise to show that these topologies are equal. Only after that can we say " $p^{-1}U$ is open in the metric topology because it is the product of 2 open sets".
$endgroup$
– DanielWainfleet
Jan 9 at 10:46
add a comment |
$begingroup$
The projection p:X×Y -> X, (x,y) -> x is continuous.
Proof. Let U be an open subset of X.
Then p$^{-1}$(U) = U×Y is open subset of X×Y
since it is the product of two open sets.
That is true of all topological spaces.
For your problem simply show that the function
d = d$_{X×Y}$ is a metric and U×Y is open when U is.
Actually, the fruitful proposition is to prove is that d
generates the product topology - whenever U is open within
X and V is open within Y, then U×V is open within X×Y.
$endgroup$
The projection p:X×Y -> X, (x,y) -> x is continuous.
Proof. Let U be an open subset of X.
Then p$^{-1}$(U) = U×Y is open subset of X×Y
since it is the product of two open sets.
That is true of all topological spaces.
For your problem simply show that the function
d = d$_{X×Y}$ is a metric and U×Y is open when U is.
Actually, the fruitful proposition is to prove is that d
generates the product topology - whenever U is open within
X and V is open within Y, then U×V is open within X×Y.
answered Jan 9 at 9:14
William ElliotWilliam Elliot
9,1962820
9,1962820
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I have shown that d is a metric. Would my original post amount to showing $U times Y$ is open given a bit of rewording? Otherwise, I cannot see how this would be done.
$endgroup$
– Andreas P
Jan 9 at 10:40
$begingroup$
The topology on $Xtimes Y$ in the Q is defined by the metric $d_{Xtimes Y}.$ Although it is true that this co-incides with the Tychonoff product topology, this is not stated in the Q, which seems to be the first part of an exercise to show that these topologies are equal. Only after that can we say " $p^{-1}U$ is open in the metric topology because it is the product of 2 open sets".
$endgroup$
– DanielWainfleet
Jan 9 at 10:46
add a comment |
$begingroup$
I have shown that d is a metric. Would my original post amount to showing $U times Y$ is open given a bit of rewording? Otherwise, I cannot see how this would be done.
$endgroup$
– Andreas P
Jan 9 at 10:40
$begingroup$
The topology on $Xtimes Y$ in the Q is defined by the metric $d_{Xtimes Y}.$ Although it is true that this co-incides with the Tychonoff product topology, this is not stated in the Q, which seems to be the first part of an exercise to show that these topologies are equal. Only after that can we say " $p^{-1}U$ is open in the metric topology because it is the product of 2 open sets".
$endgroup$
– DanielWainfleet
Jan 9 at 10:46
$begingroup$
I have shown that d is a metric. Would my original post amount to showing $U times Y$ is open given a bit of rewording? Otherwise, I cannot see how this would be done.
$endgroup$
– Andreas P
Jan 9 at 10:40
$begingroup$
I have shown that d is a metric. Would my original post amount to showing $U times Y$ is open given a bit of rewording? Otherwise, I cannot see how this would be done.
$endgroup$
– Andreas P
Jan 9 at 10:40
$begingroup$
The topology on $Xtimes Y$ in the Q is defined by the metric $d_{Xtimes Y}.$ Although it is true that this co-incides with the Tychonoff product topology, this is not stated in the Q, which seems to be the first part of an exercise to show that these topologies are equal. Only after that can we say " $p^{-1}U$ is open in the metric topology because it is the product of 2 open sets".
$endgroup$
– DanielWainfleet
Jan 9 at 10:46
$begingroup$
The topology on $Xtimes Y$ in the Q is defined by the metric $d_{Xtimes Y}.$ Although it is true that this co-incides with the Tychonoff product topology, this is not stated in the Q, which seems to be the first part of an exercise to show that these topologies are equal. Only after that can we say " $p^{-1}U$ is open in the metric topology because it is the product of 2 open sets".
$endgroup$
– DanielWainfleet
Jan 9 at 10:46
add a comment |
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