Mixing
Is every continious image of a non-compact space is non-compact?
Is every continuous image of a non-compact space is non-compact?
I was thinking about constant function. I think it will be false.
Am I right?
general-topology compactness
edited Nov 22 '18 at 13:29
Martin Sleziak
44.6k8115271
asked Nov 22 '18 at 13:07
jasminejasmine
1,645416
add a comment |
Is every continuous image of a non-compact space is non-compact?
I was thinking about constant function. I think it will be false.
Am I right?
general-topology compactness
edited Nov 22 '18 at 13:29
Martin Sleziak
44.6k8115271
asked Nov 22 '18 at 13:07
jasminejasmine
1,645416
3
If the function is constant, then the image is a singleton, which is definetely compact.
– José Carlos Santos
Nov 22 '18 at 13:10
okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
– jasmine
Nov 22 '18 at 13:17
1
Yes, the statement is false, for the reason that you gave.
– José Carlos Santos
Nov 22 '18 at 13:23
add a comment |
Is every continuous image of a non-compact space is non-compact?
I was thinking about constant function. I think it will be false.
Am I right?
general-topology compactness
edited Nov 22 '18 at 13:29
Martin Sleziak
44.6k8115271
asked Nov 22 '18 at 13:07
jasminejasmine
1,645416
Is every continuous image of a non-compact space is non-compact?
I was thinking about constant function. I think it will be false.
Am I right?
general-topology compactness
general-topology compactness
edited Nov 22 '18 at 13:29
Martin Sleziak
44.6k8115271
asked Nov 22 '18 at 13:07
jasminejasmine
1,645416
edited Nov 22 '18 at 13:29
Martin Sleziak
44.6k8115271
asked Nov 22 '18 at 13:07
jasminejasmine
1,645416
edited Nov 22 '18 at 13:29
Martin Sleziak
44.6k8115271
edited Nov 22 '18 at 13:29
Martin Sleziak
44.6k8115271
edited Nov 22 '18 at 13:29
Martin Sleziak
44.6k8115271
44.6k8115271
asked Nov 22 '18 at 13:07
jasminejasmine
1,645416
asked Nov 22 '18 at 13:07
jasminejasmine
1,645416
asked Nov 22 '18 at 13:07
jasminejasmine
1,645416
1,645416
3
If the function is constant, then the image is a singleton, which is definetely compact.
– José Carlos Santos
Nov 22 '18 at 13:10
okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
– jasmine
Nov 22 '18 at 13:17
1
Yes, the statement is false, for the reason that you gave.
– José Carlos Santos
Nov 22 '18 at 13:23
add a comment |
3
If the function is constant, then the image is a singleton, which is definetely compact.
– José Carlos Santos
Nov 22 '18 at 13:10
okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
– jasmine
Nov 22 '18 at 13:17
1
Yes, the statement is false, for the reason that you gave.
– José Carlos Santos
Nov 22 '18 at 13:23
3
3
If the function is constant, then the image is a singleton, which is definetely compact.
– José Carlos Santos
Nov 22 '18 at 13:10
If the function is constant, then the image is a singleton, which is definetely compact.
– José Carlos Santos
Nov 22 '18 at 13:10
okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
– jasmine
Nov 22 '18 at 13:17
okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
– jasmine
Nov 22 '18 at 13:17
1
1
Yes, the statement is false, for the reason that you gave.
– José Carlos Santos
Nov 22 '18 at 13:23
Yes, the statement is false, for the reason that you gave.
– José Carlos Santos
Nov 22 '18 at 13:23
add a comment |
1 Answer
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Let $f: [0, infty) to mathbb R$ be defined by
$f(x)=1-x$ , if $0 le x le 1$ and $f(x)=0$, if $x>1$.
Then $f$ is continuous, $[0, infty)$ is not compact, but $f([0, infty))=[0,1]$ is compact.
answered Nov 22 '18 at 13:20
FredFred
44.4k1845
add a comment |
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1 Answer
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Let $f: [0, infty) to mathbb R$ be defined by
$f(x)=1-x$ , if $0 le x le 1$ and $f(x)=0$, if $x>1$.
Then $f$ is continuous, $[0, infty)$ is not compact, but $f([0, infty))=[0,1]$ is compact.
answered Nov 22 '18 at 13:20
FredFred
44.4k1845
add a comment |
Let $f: [0, infty) to mathbb R$ be defined by
$f(x)=1-x$ , if $0 le x le 1$ and $f(x)=0$, if $x>1$.
Then $f$ is continuous, $[0, infty)$ is not compact, but $f([0, infty))=[0,1]$ is compact.
answered Nov 22 '18 at 13:20
FredFred
44.4k1845
add a comment |
Let $f: [0, infty) to mathbb R$ be defined by
$f(x)=1-x$ , if $0 le x le 1$ and $f(x)=0$, if $x>1$.
Then $f$ is continuous, $[0, infty)$ is not compact, but $f([0, infty))=[0,1]$ is compact.
answered Nov 22 '18 at 13:20
FredFred
44.4k1845
Let $f: [0, infty) to mathbb R$ be defined by
$f(x)=1-x$ , if $0 le x le 1$ and $f(x)=0$, if $x>1$.
Then $f$ is continuous, $[0, infty)$ is not compact, but $f([0, infty))=[0,1]$ is compact.
answered Nov 22 '18 at 13:20
FredFred
44.4k1845
answered Nov 22 '18 at 13:20
FredFred
44.4k1845
answered Nov 22 '18 at 13:20
FredFred
44.4k1845
answered Nov 22 '18 at 13:20
FredFred
44.4k1845
44.4k1845
add a comment |
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3
If the function is constant, then the image is a singleton, which is definetely compact.
– José Carlos Santos
Nov 22 '18 at 13:10
okss @ Jose carlos sir,, that mean this is the counter example,,,so statement is false .Am i right ?
– jasmine
Nov 22 '18 at 13:17
1
Yes, the statement is false, for the reason that you gave.
– José Carlos Santos
Nov 22 '18 at 13:23