Proving $f$ is continuous iff all the functions parameterized are continuous
$begingroup$
Let $(Y,tau)$ and $(X_i,tau_i),i=1,2,...n$ be the topological spaces. Further for each i , let $f_i$ be a mapping of $(Y,tau)$ into $(X_i,tau_i)$. Porve that the mapping $f:(Y,tau)toprod_limits{i=1}^{n}(X_i,tau_i)$, given by $f(y)=(f_1(y),f_2(y),...,f_n(y))$ is continuous if and only if every $f_i$ is continuous.
My proof:
$rightarrow$ Let $U$ be an open set of $(prod_limits{i=1}^{n}X_i,tau_i)$
Suppose $f$ is continuous then $f^{-1}(U)intau$
$f_i(U)=p_icirc f(U)$ where $p_i$ is the projection $p_i:(prod_limits{i=1}^{n}X_i,tau_i)to (X_itau_i)$.
As $f_i(U)$ is a composition of two continuous functions hence it is continuous.
$leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n$ be an open set in the subspace topology $f(Y),tau_{nY}$
Supposing $f_i$ is continuous forall the $i=1,2...n$ then
$f^{-1}_icirc p_i(U)=f^{-1}(U_i)intau_i$
But $f^{-1}(U)=f^{-1}_icirc p_i(U)$ then $f$ is continuous.
Question:
Is my proof right? If not. Why not?
Thanks in advance!
general-topology proof-verification
asked Dec 24 '18 at 15:10
Pedro GomesPedro Gomes
2,0162823
$endgroup$
add a comment |
$begingroup$
Let $(Y,tau)$ and $(X_i,tau_i),i=1,2,...n$ be the topological spaces. Further for each i , let $f_i$ be a mapping of $(Y,tau)$ into $(X_i,tau_i)$. Porve that the mapping $f:(Y,tau)toprod_limits{i=1}^{n}(X_i,tau_i)$, given by $f(y)=(f_1(y),f_2(y),...,f_n(y))$ is continuous if and only if every $f_i$ is continuous.
My proof:
$rightarrow$ Let $U$ be an open set of $(prod_limits{i=1}^{n}X_i,tau_i)$
Suppose $f$ is continuous then $f^{-1}(U)intau$
$f_i(U)=p_icirc f(U)$ where $p_i$ is the projection $p_i:(prod_limits{i=1}^{n}X_i,tau_i)to (X_itau_i)$.
As $f_i(U)$ is a composition of two continuous functions hence it is continuous.
$leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n$ be an open set in the subspace topology $f(Y),tau_{nY}$
Supposing $f_i$ is continuous forall the $i=1,2...n$ then
$f^{-1}_icirc p_i(U)=f^{-1}(U_i)intau_i$
But $f^{-1}(U)=f^{-1}_icirc p_i(U)$ then $f$ is continuous.
Question:
Is my proof right? If not. Why not?
Thanks in advance!
general-topology proof-verification
asked Dec 24 '18 at 15:10
Pedro GomesPedro Gomes
2,0162823
$endgroup$
add a comment |
$begingroup$
Let $(Y,tau)$ and $(X_i,tau_i),i=1,2,...n$ be the topological spaces. Further for each i , let $f_i$ be a mapping of $(Y,tau)$ into $(X_i,tau_i)$. Porve that the mapping $f:(Y,tau)toprod_limits{i=1}^{n}(X_i,tau_i)$, given by $f(y)=(f_1(y),f_2(y),...,f_n(y))$ is continuous if and only if every $f_i$ is continuous.
My proof:
$rightarrow$ Let $U$ be an open set of $(prod_limits{i=1}^{n}X_i,tau_i)$
Suppose $f$ is continuous then $f^{-1}(U)intau$
$f_i(U)=p_icirc f(U)$ where $p_i$ is the projection $p_i:(prod_limits{i=1}^{n}X_i,tau_i)to (X_itau_i)$.
As $f_i(U)$ is a composition of two continuous functions hence it is continuous.
$leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n$ be an open set in the subspace topology $f(Y),tau_{nY}$
Supposing $f_i$ is continuous forall the $i=1,2...n$ then
$f^{-1}_icirc p_i(U)=f^{-1}(U_i)intau_i$
But $f^{-1}(U)=f^{-1}_icirc p_i(U)$ then $f$ is continuous.
Question:
Is my proof right? If not. Why not?
Thanks in advance!
general-topology proof-verification
asked Dec 24 '18 at 15:10
Pedro GomesPedro Gomes
2,0162823
$endgroup$
Let $(Y,tau)$ and $(X_i,tau_i),i=1,2,...n$ be the topological spaces. Further for each i , let $f_i$ be a mapping of $(Y,tau)$ into $(X_i,tau_i)$. Porve that the mapping $f:(Y,tau)toprod_limits{i=1}^{n}(X_i,tau_i)$, given by $f(y)=(f_1(y),f_2(y),...,f_n(y))$ is continuous if and only if every $f_i$ is continuous.
My proof:
$rightarrow$ Let $U$ be an open set of $(prod_limits{i=1}^{n}X_i,tau_i)$
Suppose $f$ is continuous then $f^{-1}(U)intau$
$f_i(U)=p_icirc f(U)$ where $p_i$ is the projection $p_i:(prod_limits{i=1}^{n}X_i,tau_i)to (X_itau_i)$.
As $f_i(U)$ is a composition of two continuous functions hence it is continuous.
$leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n$ be an open set in the subspace topology $f(Y),tau_{nY}$
Supposing $f_i$ is continuous forall the $i=1,2...n$ then
$f^{-1}_icirc p_i(U)=f^{-1}(U_i)intau_i$
But $f^{-1}(U)=f^{-1}_icirc p_i(U)$ then $f$ is continuous.
Question:
Is my proof right? If not. Why not?
Thanks in advance!
general-topology proof-verification
general-topology proof-verification
asked Dec 24 '18 at 15:10
Pedro GomesPedro Gomes
2,0162823
asked Dec 24 '18 at 15:10
Pedro GomesPedro Gomes
2,0162823
asked Dec 24 '18 at 15:10
Pedro GomesPedro Gomes
2,0162823
asked Dec 24 '18 at 15:10
Pedro GomesPedro Gomes
2,0162823
asked Dec 24 '18 at 15:10
Pedro GomesPedro Gomes
2,0162823
2,0162823
add a comment |
add a comment |
1 Answer
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I would recommend the following change
$leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n subset prod_limits{i=1}^{n}(X_i,tau_i)$ with each $U_i$ open $X_i$, so that $U$ is open in the product space. These open sets form a basis, so we only have to show that $f^{-1}(U)$ must be open in $Y$ to prove the continuity of $f$.
answered Dec 24 '18 at 16:08
CopyPasteItCopyPasteIt
4,4321828
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I would recommend the following change
$leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n subset prod_limits{i=1}^{n}(X_i,tau_i)$ with each $U_i$ open $X_i$, so that $U$ is open in the product space. These open sets form a basis, so we only have to show that $f^{-1}(U)$ must be open in $Y$ to prove the continuity of $f$.
answered Dec 24 '18 at 16:08
CopyPasteItCopyPasteIt
4,4321828
$endgroup$
add a comment |
$begingroup$
I would recommend the following change
$leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n subset prod_limits{i=1}^{n}(X_i,tau_i)$ with each $U_i$ open $X_i$, so that $U$ is open in the product space. These open sets form a basis, so we only have to show that $f^{-1}(U)$ must be open in $Y$ to prove the continuity of $f$.
answered Dec 24 '18 at 16:08
CopyPasteItCopyPasteIt
4,4321828
$endgroup$
add a comment |
$begingroup$
I would recommend the following change
$leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n subset prod_limits{i=1}^{n}(X_i,tau_i)$ with each $U_i$ open $X_i$, so that $U$ is open in the product space. These open sets form a basis, so we only have to show that $f^{-1}(U)$ must be open in $Y$ to prove the continuity of $f$.
answered Dec 24 '18 at 16:08
CopyPasteItCopyPasteIt
4,4321828
$endgroup$
I would recommend the following change
$leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n subset prod_limits{i=1}^{n}(X_i,tau_i)$ with each $U_i$ open $X_i$, so that $U$ is open in the product space. These open sets form a basis, so we only have to show that $f^{-1}(U)$ must be open in $Y$ to prove the continuity of $f$.
answered Dec 24 '18 at 16:08
CopyPasteItCopyPasteIt
4,4321828
answered Dec 24 '18 at 16:08
CopyPasteItCopyPasteIt
4,4321828
answered Dec 24 '18 at 16:08
CopyPasteItCopyPasteIt
4,4321828
answered Dec 24 '18 at 16:08
CopyPasteItCopyPasteIt
4,4321828
4,4321828
add a comment |
add a comment |
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