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The Domain of a continuous function is an open set
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I have started studying the book "Differentiable manifolds; An Introduction" by Brickell & Clark. But I have encountered the following paragraph on page 4.
Why is the domain of a continuous functions nacessarily an open set?
Thanks for any help.
geometry metric-spaces differential-topology
asked Jan 18 at 7:59
useruser
623311
$endgroup$
add a comment |
$begingroup$
I have started studying the book "Differentiable manifolds; An Introduction" by Brickell & Clark. But I have encountered the following paragraph on page 4.
Why is the domain of a continuous functions nacessarily an open set?
Thanks for any help.
geometry metric-spaces differential-topology
asked Jan 18 at 7:59
useruser
623311
$endgroup$
$begingroup$
If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:10
add a comment |
$begingroup$
I have started studying the book "Differentiable manifolds; An Introduction" by Brickell & Clark. But I have encountered the following paragraph on page 4.
Why is the domain of a continuous functions nacessarily an open set?
Thanks for any help.
geometry metric-spaces differential-topology
asked Jan 18 at 7:59
useruser
623311
$endgroup$
I have started studying the book "Differentiable manifolds; An Introduction" by Brickell & Clark. But I have encountered the following paragraph on page 4.
Why is the domain of a continuous functions nacessarily an open set?
Thanks for any help.
geometry metric-spaces differential-topology
geometry metric-spaces differential-topology
asked Jan 18 at 7:59
useruser
623311
asked Jan 18 at 7:59
useruser
623311
edited Jan 18 at 10:11
user
asked Jan 18 at 7:59
useruser
623311
asked Jan 18 at 7:59
useruser
623311
asked Jan 18 at 7:59
useruser
623311
623311
$begingroup$
If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:10
add a comment |
$begingroup$
If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:10
$begingroup$
If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:10
$begingroup$
If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:10
add a comment |
2 Answers
2
active
oldest
votes
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Let $X,Y$ be topological spaces. Then by topological definition $f:X rightarrow Y$ is a continuous map iff preimage $f^{-1}(U)$ of any open set $U subset Y$ is open. The whole $Y$ itself is an open set by definition of a topological space. Since $f^{-1}(Y)=X$ we get that $X$ is open for continuous $f$.
Essense. In the topological definition of a continuous map $f$ we forget that the domain of $f$ is a subset of some space and treat it as a whole space, and in this space it is indeed an open set.
answered Jan 18 at 8:25
ZeeklessZeekless
577111
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add a comment |
$begingroup$
$S$ and $T$ are topological spaces and the domain of $f:S to T$ is the set $S$ and $S$ is open.
answered Jan 18 at 8:09
FredFred
46.9k1848
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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$begingroup$
Let $X,Y$ be topological spaces. Then by topological definition $f:X rightarrow Y$ is a continuous map iff preimage $f^{-1}(U)$ of any open set $U subset Y$ is open. The whole $Y$ itself is an open set by definition of a topological space. Since $f^{-1}(Y)=X$ we get that $X$ is open for continuous $f$.
Essense. In the topological definition of a continuous map $f$ we forget that the domain of $f$ is a subset of some space and treat it as a whole space, and in this space it is indeed an open set.
answered Jan 18 at 8:25
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be topological spaces. Then by topological definition $f:X rightarrow Y$ is a continuous map iff preimage $f^{-1}(U)$ of any open set $U subset Y$ is open. The whole $Y$ itself is an open set by definition of a topological space. Since $f^{-1}(Y)=X$ we get that $X$ is open for continuous $f$.
Essense. In the topological definition of a continuous map $f$ we forget that the domain of $f$ is a subset of some space and treat it as a whole space, and in this space it is indeed an open set.
answered Jan 18 at 8:25
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be topological spaces. Then by topological definition $f:X rightarrow Y$ is a continuous map iff preimage $f^{-1}(U)$ of any open set $U subset Y$ is open. The whole $Y$ itself is an open set by definition of a topological space. Since $f^{-1}(Y)=X$ we get that $X$ is open for continuous $f$.
Essense. In the topological definition of a continuous map $f$ we forget that the domain of $f$ is a subset of some space and treat it as a whole space, and in this space it is indeed an open set.
answered Jan 18 at 8:25
ZeeklessZeekless
577111
$endgroup$
Let $X,Y$ be topological spaces. Then by topological definition $f:X rightarrow Y$ is a continuous map iff preimage $f^{-1}(U)$ of any open set $U subset Y$ is open. The whole $Y$ itself is an open set by definition of a topological space. Since $f^{-1}(Y)=X$ we get that $X$ is open for continuous $f$.
Essense. In the topological definition of a continuous map $f$ we forget that the domain of $f$ is a subset of some space and treat it as a whole space, and in this space it is indeed an open set.
answered Jan 18 at 8:25
ZeeklessZeekless
577111
answered Jan 18 at 8:25
ZeeklessZeekless
577111
answered Jan 18 at 8:25
ZeeklessZeekless
577111
answered Jan 18 at 8:25
ZeeklessZeekless
577111
577111
add a comment |
add a comment |
$begingroup$
$S$ and $T$ are topological spaces and the domain of $f:S to T$ is the set $S$ and $S$ is open.
answered Jan 18 at 8:09
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
$S$ and $T$ are topological spaces and the domain of $f:S to T$ is the set $S$ and $S$ is open.
answered Jan 18 at 8:09
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
$S$ and $T$ are topological spaces and the domain of $f:S to T$ is the set $S$ and $S$ is open.
answered Jan 18 at 8:09
FredFred
46.9k1848
$endgroup$
$S$ and $T$ are topological spaces and the domain of $f:S to T$ is the set $S$ and $S$ is open.
answered Jan 18 at 8:09
FredFred
46.9k1848
answered Jan 18 at 8:09
FredFred
46.9k1848
answered Jan 18 at 8:09
FredFred
46.9k1848
answered Jan 18 at 8:09
FredFred
46.9k1848
46.9k1848
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add a comment |
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$begingroup$
If the author wants to restrict the definition to open sets that is his choice. But continuous function on sets other than open sets are very much in use in Mathematics.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:10