Example of closed and discontinuous transformation
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Give an example of a closed and discontinuous transformation. This task is from compact spaces section.
I have problem with this, i can't find any solution.
general-topology
asked Jan 8 at 11:45
VictorVictor
464
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add a comment |
$begingroup$
Give an example of a closed and discontinuous transformation. This task is from compact spaces section.
I have problem with this, i can't find any solution.
general-topology
asked Jan 8 at 11:45
VictorVictor
464
$endgroup$
add a comment |
$begingroup$
Give an example of a closed and discontinuous transformation. This task is from compact spaces section.
I have problem with this, i can't find any solution.
general-topology
asked Jan 8 at 11:45
VictorVictor
464
$endgroup$
Give an example of a closed and discontinuous transformation. This task is from compact spaces section.
I have problem with this, i can't find any solution.
general-topology
general-topology
asked Jan 8 at 11:45
VictorVictor
464
asked Jan 8 at 11:45
VictorVictor
464
asked Jan 8 at 11:45
VictorVictor
464
asked Jan 8 at 11:45
VictorVictor
464
asked Jan 8 at 11:45
VictorVictor
464
464
add a comment |
add a comment |
1 Answer
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Let $f$ be the identity map from $mathbb R$ with usual metric to $mathbb R$ with discrete metric. Then $f$ is not continuous but it is a closed map. Another example where both spaces are compact: define $f:[0,1] to {0,1}$ by $f(x)=0$ if $x <frac 1 2$ and $f(x)=1$ otherwise. Note that image of any set is closed!.
answered Jan 8 at 11:56
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
Okay, but when f: X -> Y, where X is usual metric on R and Y is discrete metric on R is not continuous? Because closed is clear i think (Every set in discrete metric is closed and open).
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– Victor
Jan 8 at 12:26
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@Victor It is not continuous. Every set is open in $Y$ and if the identity map is continuous then every set would be open in usual topology which is not true. See also my second example.
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– Kavi Rama Murthy
Jan 8 at 12:32
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
Let $f$ be the identity map from $mathbb R$ with usual metric to $mathbb R$ with discrete metric. Then $f$ is not continuous but it is a closed map. Another example where both spaces are compact: define $f:[0,1] to {0,1}$ by $f(x)=0$ if $x <frac 1 2$ and $f(x)=1$ otherwise. Note that image of any set is closed!.
answered Jan 8 at 11:56
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
Okay, but when f: X -> Y, where X is usual metric on R and Y is discrete metric on R is not continuous? Because closed is clear i think (Every set in discrete metric is closed and open).
$endgroup$
– Victor
Jan 8 at 12:26
$begingroup$
@Victor It is not continuous. Every set is open in $Y$ and if the identity map is continuous then every set would be open in usual topology which is not true. See also my second example.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 12:32
add a comment |
$begingroup$
Let $f$ be the identity map from $mathbb R$ with usual metric to $mathbb R$ with discrete metric. Then $f$ is not continuous but it is a closed map. Another example where both spaces are compact: define $f:[0,1] to {0,1}$ by $f(x)=0$ if $x <frac 1 2$ and $f(x)=1$ otherwise. Note that image of any set is closed!.
answered Jan 8 at 11:56
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
Okay, but when f: X -> Y, where X is usual metric on R and Y is discrete metric on R is not continuous? Because closed is clear i think (Every set in discrete metric is closed and open).
$endgroup$
– Victor
Jan 8 at 12:26
$begingroup$
@Victor It is not continuous. Every set is open in $Y$ and if the identity map is continuous then every set would be open in usual topology which is not true. See also my second example.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 12:32
add a comment |
$begingroup$
Let $f$ be the identity map from $mathbb R$ with usual metric to $mathbb R$ with discrete metric. Then $f$ is not continuous but it is a closed map. Another example where both spaces are compact: define $f:[0,1] to {0,1}$ by $f(x)=0$ if $x <frac 1 2$ and $f(x)=1$ otherwise. Note that image of any set is closed!.
answered Jan 8 at 11:56
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
Let $f$ be the identity map from $mathbb R$ with usual metric to $mathbb R$ with discrete metric. Then $f$ is not continuous but it is a closed map. Another example where both spaces are compact: define $f:[0,1] to {0,1}$ by $f(x)=0$ if $x <frac 1 2$ and $f(x)=1$ otherwise. Note that image of any set is closed!.
answered Jan 8 at 11:56
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
edited Jan 8 at 12:34
answered Jan 8 at 11:56
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Jan 8 at 11:56
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Jan 8 at 11:56
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
74.6k53270
$begingroup$
Okay, but when f: X -> Y, where X is usual metric on R and Y is discrete metric on R is not continuous? Because closed is clear i think (Every set in discrete metric is closed and open).
$endgroup$
– Victor
Jan 8 at 12:26
$begingroup$
@Victor It is not continuous. Every set is open in $Y$ and if the identity map is continuous then every set would be open in usual topology which is not true. See also my second example.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 12:32
add a comment |
$begingroup$
Okay, but when f: X -> Y, where X is usual metric on R and Y is discrete metric on R is not continuous? Because closed is clear i think (Every set in discrete metric is closed and open).
$endgroup$
– Victor
Jan 8 at 12:26
$begingroup$
@Victor It is not continuous. Every set is open in $Y$ and if the identity map is continuous then every set would be open in usual topology which is not true. See also my second example.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 12:32
$begingroup$
Okay, but when f: X -> Y, where X is usual metric on R and Y is discrete metric on R is not continuous? Because closed is clear i think (Every set in discrete metric is closed and open).
$endgroup$
– Victor
Jan 8 at 12:26
$begingroup$
Okay, but when f: X -> Y, where X is usual metric on R and Y is discrete metric on R is not continuous? Because closed is clear i think (Every set in discrete metric is closed and open).
$endgroup$
– Victor
Jan 8 at 12:26
$begingroup$
@Victor It is not continuous. Every set is open in $Y$ and if the identity map is continuous then every set would be open in usual topology which is not true. See also my second example.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 12:32
$begingroup$
@Victor It is not continuous. Every set is open in $Y$ and if the identity map is continuous then every set would be open in usual topology which is not true. See also my second example.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 12:32
add a comment |
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