Infinite set with cofinite topology
$begingroup$
Prove: If $X$ is an infinite set with the cofinite topology, then every $f in C(X)$ is constant.
Attempt: I can show that $X$, an infinite set with the cofinite topology is connected. Are all continuous functions with connected, cofininte topology constant? If so, how? Or is there another way to solve the original problem.This is from Folland's Real Analysis book.
real-analysis general-topology
asked Nov 25 '18 at 12:25
DomDom
1627
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add a comment |
$begingroup$
Prove: If $X$ is an infinite set with the cofinite topology, then every $f in C(X)$ is constant.
Attempt: I can show that $X$, an infinite set with the cofinite topology is connected. Are all continuous functions with connected, cofininte topology constant? If so, how? Or is there another way to solve the original problem.This is from Folland's Real Analysis book.
real-analysis general-topology
asked Nov 25 '18 at 12:25
DomDom
1627
$endgroup$
add a comment |
$begingroup$
Prove: If $X$ is an infinite set with the cofinite topology, then every $f in C(X)$ is constant.
Attempt: I can show that $X$, an infinite set with the cofinite topology is connected. Are all continuous functions with connected, cofininte topology constant? If so, how? Or is there another way to solve the original problem.This is from Folland's Real Analysis book.
real-analysis general-topology
asked Nov 25 '18 at 12:25
DomDom
1627
$endgroup$
Prove: If $X$ is an infinite set with the cofinite topology, then every $f in C(X)$ is constant.
Attempt: I can show that $X$, an infinite set with the cofinite topology is connected. Are all continuous functions with connected, cofininte topology constant? If so, how? Or is there another way to solve the original problem.This is from Folland's Real Analysis book.
real-analysis general-topology
real-analysis general-topology
asked Nov 25 '18 at 12:25
DomDom
1627
asked Nov 25 '18 at 12:25
DomDom
1627
asked Nov 25 '18 at 12:25
DomDom
1627
asked Nov 25 '18 at 12:25
DomDom
1627
asked Nov 25 '18 at 12:25
DomDom
1627
1627
add a comment |
add a comment |
1 Answer
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$begingroup$
If $f$ is not constant it has at least two values, say $p$ and $q$ in the reals.
We can find two disjoint open intervals $I=(p-r,p+r)$ and $I'=(q-r,q+r)$ in the reals that are disjoint (for small enough $r$), and then by continuity of $f$, both $O=f^{-1}[I]$ and $O' = f^{-1}[I']$ are disjoint non-empty (as $p$ and $q$ are assumed values) open sets in $X$.
But such open sets cannot exist: $O=Xsetminus F$ for some finite set $F$ and $O' = Xsetminus F'$ for some finite set $F'$. Any point in $X setminus (F cup F')$, which must exist as the union of two finite sets is finite and $X$ is infinite, is in both $O$ and $O'$, so non-empty open sets in $X$ (in the cofinite topology) always intersect. This property is called hyperconnected, FYI.
So any continuous map from a hyperconnected space to a Hausdorff space is constant, is what the previous proof shows.
answered Nov 25 '18 at 12:38
Henno BrandsmaHenno Brandsma
106k347114
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1 Answer
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$begingroup$
If $f$ is not constant it has at least two values, say $p$ and $q$ in the reals.
We can find two disjoint open intervals $I=(p-r,p+r)$ and $I'=(q-r,q+r)$ in the reals that are disjoint (for small enough $r$), and then by continuity of $f$, both $O=f^{-1}[I]$ and $O' = f^{-1}[I']$ are disjoint non-empty (as $p$ and $q$ are assumed values) open sets in $X$.
But such open sets cannot exist: $O=Xsetminus F$ for some finite set $F$ and $O' = Xsetminus F'$ for some finite set $F'$. Any point in $X setminus (F cup F')$, which must exist as the union of two finite sets is finite and $X$ is infinite, is in both $O$ and $O'$, so non-empty open sets in $X$ (in the cofinite topology) always intersect. This property is called hyperconnected, FYI.
So any continuous map from a hyperconnected space to a Hausdorff space is constant, is what the previous proof shows.
answered Nov 25 '18 at 12:38
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
If $f$ is not constant it has at least two values, say $p$ and $q$ in the reals.
We can find two disjoint open intervals $I=(p-r,p+r)$ and $I'=(q-r,q+r)$ in the reals that are disjoint (for small enough $r$), and then by continuity of $f$, both $O=f^{-1}[I]$ and $O' = f^{-1}[I']$ are disjoint non-empty (as $p$ and $q$ are assumed values) open sets in $X$.
But such open sets cannot exist: $O=Xsetminus F$ for some finite set $F$ and $O' = Xsetminus F'$ for some finite set $F'$. Any point in $X setminus (F cup F')$, which must exist as the union of two finite sets is finite and $X$ is infinite, is in both $O$ and $O'$, so non-empty open sets in $X$ (in the cofinite topology) always intersect. This property is called hyperconnected, FYI.
So any continuous map from a hyperconnected space to a Hausdorff space is constant, is what the previous proof shows.
answered Nov 25 '18 at 12:38
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
If $f$ is not constant it has at least two values, say $p$ and $q$ in the reals.
We can find two disjoint open intervals $I=(p-r,p+r)$ and $I'=(q-r,q+r)$ in the reals that are disjoint (for small enough $r$), and then by continuity of $f$, both $O=f^{-1}[I]$ and $O' = f^{-1}[I']$ are disjoint non-empty (as $p$ and $q$ are assumed values) open sets in $X$.
But such open sets cannot exist: $O=Xsetminus F$ for some finite set $F$ and $O' = Xsetminus F'$ for some finite set $F'$. Any point in $X setminus (F cup F')$, which must exist as the union of two finite sets is finite and $X$ is infinite, is in both $O$ and $O'$, so non-empty open sets in $X$ (in the cofinite topology) always intersect. This property is called hyperconnected, FYI.
So any continuous map from a hyperconnected space to a Hausdorff space is constant, is what the previous proof shows.
answered Nov 25 '18 at 12:38
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
If $f$ is not constant it has at least two values, say $p$ and $q$ in the reals.
We can find two disjoint open intervals $I=(p-r,p+r)$ and $I'=(q-r,q+r)$ in the reals that are disjoint (for small enough $r$), and then by continuity of $f$, both $O=f^{-1}[I]$ and $O' = f^{-1}[I']$ are disjoint non-empty (as $p$ and $q$ are assumed values) open sets in $X$.
But such open sets cannot exist: $O=Xsetminus F$ for some finite set $F$ and $O' = Xsetminus F'$ for some finite set $F'$. Any point in $X setminus (F cup F')$, which must exist as the union of two finite sets is finite and $X$ is infinite, is in both $O$ and $O'$, so non-empty open sets in $X$ (in the cofinite topology) always intersect. This property is called hyperconnected, FYI.
So any continuous map from a hyperconnected space to a Hausdorff space is constant, is what the previous proof shows.
answered Nov 25 '18 at 12:38
Henno BrandsmaHenno Brandsma
106k347114
answered Nov 25 '18 at 12:38
Henno BrandsmaHenno Brandsma
106k347114
answered Nov 25 '18 at 12:38
Henno BrandsmaHenno Brandsma
106k347114
answered Nov 25 '18 at 12:38
Henno BrandsmaHenno Brandsma
106k347114
106k347114
add a comment |
add a comment |
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