Continuous extension of a function to the Stone-Cech compactification
$begingroup$
Let $C(X)$ be the set of all continuous, real-valued functions on a Tychonoff (completely regular)
topological space $X$ and let $beta X$ be the Stone-Cech compactification of $X$. Now let $fin C(X)$ such that $f(X)={0, 1}$ and $f^beta$ be an extension of $f$ to $beta X$, that is, $f^betain C(beta X)$. How can we show that $f^beta(beta X)={0, 1}$?
general-topology compactification
edited Dec 28 '18 at 13:54
F M
3,09652341
asked Dec 25 '18 at 10:22
Andy.GAndy.G
935
$endgroup$
add a comment |
$begingroup$
Let $C(X)$ be the set of all continuous, real-valued functions on a Tychonoff (completely regular)
topological space $X$ and let $beta X$ be the Stone-Cech compactification of $X$. Now let $fin C(X)$ such that $f(X)={0, 1}$ and $f^beta$ be an extension of $f$ to $beta X$, that is, $f^betain C(beta X)$. How can we show that $f^beta(beta X)={0, 1}$?
general-topology compactification
edited Dec 28 '18 at 13:54
F M
3,09652341
asked Dec 25 '18 at 10:22
Andy.GAndy.G
935
$endgroup$
add a comment |
$begingroup$
Let $C(X)$ be the set of all continuous, real-valued functions on a Tychonoff (completely regular)
topological space $X$ and let $beta X$ be the Stone-Cech compactification of $X$. Now let $fin C(X)$ such that $f(X)={0, 1}$ and $f^beta$ be an extension of $f$ to $beta X$, that is, $f^betain C(beta X)$. How can we show that $f^beta(beta X)={0, 1}$?
general-topology compactification
edited Dec 28 '18 at 13:54
F M
3,09652341
asked Dec 25 '18 at 10:22
Andy.GAndy.G
935
$endgroup$
Let $C(X)$ be the set of all continuous, real-valued functions on a Tychonoff (completely regular)
topological space $X$ and let $beta X$ be the Stone-Cech compactification of $X$. Now let $fin C(X)$ such that $f(X)={0, 1}$ and $f^beta$ be an extension of $f$ to $beta X$, that is, $f^betain C(beta X)$. How can we show that $f^beta(beta X)={0, 1}$?
general-topology compactification
general-topology compactification
edited Dec 28 '18 at 13:54
F M
3,09652341
asked Dec 25 '18 at 10:22
Andy.GAndy.G
935
edited Dec 28 '18 at 13:54
F M
3,09652341
asked Dec 25 '18 at 10:22
Andy.GAndy.G
935
edited Dec 28 '18 at 13:54
F M
3,09652341
edited Dec 28 '18 at 13:54
F M
3,09652341
edited Dec 28 '18 at 13:54
F M
3,09652341
3,09652341
asked Dec 25 '18 at 10:22
Andy.GAndy.G
935
asked Dec 25 '18 at 10:22
Andy.GAndy.G
935
asked Dec 25 '18 at 10:22
Andy.GAndy.G
935
935
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The Stone-Čech compactification has a very strong property: if $fcolon Xto Y$ is a continuous map with $Y$ compact, then $f$ can be (uniquely) extended to $beta fcolonbeta Xto Y$. See Universal property and functoriality on Wikipedia.
Since ${0,1}$ is compact, you're done.
answered Dec 25 '18 at 11:12
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
$X$ is dense in $beta(X)$. Hence $f^{beta} (beta(X)) = f^{beta} (overline {X}) $ which is contained in the closure of ${0,1}$ (by continuity) so it is contained in ${0,1}$. Reverse inclusion follows by the fact that $f^{beta}$ extends $f$.
answered Dec 25 '18 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The Stone-Čech compactification has a very strong property: if $fcolon Xto Y$ is a continuous map with $Y$ compact, then $f$ can be (uniquely) extended to $beta fcolonbeta Xto Y$. See Universal property and functoriality on Wikipedia.
Since ${0,1}$ is compact, you're done.
answered Dec 25 '18 at 11:12
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
The Stone-Čech compactification has a very strong property: if $fcolon Xto Y$ is a continuous map with $Y$ compact, then $f$ can be (uniquely) extended to $beta fcolonbeta Xto Y$. See Universal property and functoriality on Wikipedia.
Since ${0,1}$ is compact, you're done.
answered Dec 25 '18 at 11:12
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
The Stone-Čech compactification has a very strong property: if $fcolon Xto Y$ is a continuous map with $Y$ compact, then $f$ can be (uniquely) extended to $beta fcolonbeta Xto Y$. See Universal property and functoriality on Wikipedia.
Since ${0,1}$ is compact, you're done.
answered Dec 25 '18 at 11:12
egregegreg
185k1486208
$endgroup$
The Stone-Čech compactification has a very strong property: if $fcolon Xto Y$ is a continuous map with $Y$ compact, then $f$ can be (uniquely) extended to $beta fcolonbeta Xto Y$. See Universal property and functoriality on Wikipedia.
Since ${0,1}$ is compact, you're done.
answered Dec 25 '18 at 11:12
egregegreg
185k1486208
answered Dec 25 '18 at 11:12
egregegreg
185k1486208
answered Dec 25 '18 at 11:12
egregegreg
185k1486208
answered Dec 25 '18 at 11:12
egregegreg
185k1486208
185k1486208
add a comment |
add a comment |
$begingroup$
$X$ is dense in $beta(X)$. Hence $f^{beta} (beta(X)) = f^{beta} (overline {X}) $ which is contained in the closure of ${0,1}$ (by continuity) so it is contained in ${0,1}$. Reverse inclusion follows by the fact that $f^{beta}$ extends $f$.
answered Dec 25 '18 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
$endgroup$
add a comment |
$begingroup$
$X$ is dense in $beta(X)$. Hence $f^{beta} (beta(X)) = f^{beta} (overline {X}) $ which is contained in the closure of ${0,1}$ (by continuity) so it is contained in ${0,1}$. Reverse inclusion follows by the fact that $f^{beta}$ extends $f$.
answered Dec 25 '18 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
$endgroup$
add a comment |
$begingroup$
$X$ is dense in $beta(X)$. Hence $f^{beta} (beta(X)) = f^{beta} (overline {X}) $ which is contained in the closure of ${0,1}$ (by continuity) so it is contained in ${0,1}$. Reverse inclusion follows by the fact that $f^{beta}$ extends $f$.
answered Dec 25 '18 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
$endgroup$
$X$ is dense in $beta(X)$. Hence $f^{beta} (beta(X)) = f^{beta} (overline {X}) $ which is contained in the closure of ${0,1}$ (by continuity) so it is contained in ${0,1}$. Reverse inclusion follows by the fact that $f^{beta}$ extends $f$.
answered Dec 25 '18 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
edited Dec 25 '18 at 10:36
answered Dec 25 '18 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
answered Dec 25 '18 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
answered Dec 25 '18 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
72.9k53170
add a comment |
add a comment |
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