True or false( invertibility of $f$)
$begingroup$
Let $f$ : $(1,10)$ $to$ $[2,11]$
be a continuous function, then , $f$ cannot be an invertible function.
True/false
Both domain and codomain has same cardinality so we can find a bijection between both sets.
But, the statement is true. What property is used in this problem?
calculus
asked Dec 7 '18 at 8:45
MathsaddictMathsaddict
2998
$endgroup$
add a comment |
$begingroup$
Let $f$ : $(1,10)$ $to$ $[2,11]$
be a continuous function, then , $f$ cannot be an invertible function.
True/false
Both domain and codomain has same cardinality so we can find a bijection between both sets.
But, the statement is true. What property is used in this problem?
calculus
asked Dec 7 '18 at 8:45
MathsaddictMathsaddict
2998
$endgroup$
add a comment |
$begingroup$
Let $f$ : $(1,10)$ $to$ $[2,11]$
be a continuous function, then , $f$ cannot be an invertible function.
True/false
Both domain and codomain has same cardinality so we can find a bijection between both sets.
But, the statement is true. What property is used in this problem?
calculus
asked Dec 7 '18 at 8:45
MathsaddictMathsaddict
2998
$endgroup$
Let $f$ : $(1,10)$ $to$ $[2,11]$
be a continuous function, then , $f$ cannot be an invertible function.
True/false
Both domain and codomain has same cardinality so we can find a bijection between both sets.
But, the statement is true. What property is used in this problem?
calculus
calculus
asked Dec 7 '18 at 8:45
MathsaddictMathsaddict
2998
asked Dec 7 '18 at 8:45
MathsaddictMathsaddict
2998
edited Dec 7 '18 at 9:55
Mathsaddict
asked Dec 7 '18 at 8:45
MathsaddictMathsaddict
2998
asked Dec 7 '18 at 8:45
MathsaddictMathsaddict
2998
asked Dec 7 '18 at 8:45
MathsaddictMathsaddict
2998
2998
add a comment |
add a comment |
1 Answer
1
active
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$begingroup$
If it is a bijection then the inverse would be continuous too and range of the inverse, which is $(0,10)$ would be compact because continuous image of compact sets is compact, This is a contradiction. [A continuous bijection is strictly monotonic and this makes the inverse also monotonic and continuous].
answered Dec 7 '18 at 8:50
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
$begingroup$
@WilliamSun Continuous bijections on the real line are known to have continuous inverses automatically. I am assuming this result with a hint that this can b e proved using monotonicity.
$endgroup$
– Kavi Rama Murthy
Dec 7 '18 at 8:55
$begingroup$
It was very simple to under.stand. Thanks.
$endgroup$
– Mathsaddict
Dec 7 '18 at 9:04
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
If it is a bijection then the inverse would be continuous too and range of the inverse, which is $(0,10)$ would be compact because continuous image of compact sets is compact, This is a contradiction. [A continuous bijection is strictly monotonic and this makes the inverse also monotonic and continuous].
answered Dec 7 '18 at 8:50
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
$begingroup$
@WilliamSun Continuous bijections on the real line are known to have continuous inverses automatically. I am assuming this result with a hint that this can b e proved using monotonicity.
$endgroup$
– Kavi Rama Murthy
Dec 7 '18 at 8:55
$begingroup$
It was very simple to under.stand. Thanks.
$endgroup$
– Mathsaddict
Dec 7 '18 at 9:04
add a comment |
$begingroup$
If it is a bijection then the inverse would be continuous too and range of the inverse, which is $(0,10)$ would be compact because continuous image of compact sets is compact, This is a contradiction. [A continuous bijection is strictly monotonic and this makes the inverse also monotonic and continuous].
answered Dec 7 '18 at 8:50
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
$begingroup$
@WilliamSun Continuous bijections on the real line are known to have continuous inverses automatically. I am assuming this result with a hint that this can b e proved using monotonicity.
$endgroup$
– Kavi Rama Murthy
Dec 7 '18 at 8:55
$begingroup$
It was very simple to under.stand. Thanks.
$endgroup$
– Mathsaddict
Dec 7 '18 at 9:04
add a comment |
$begingroup$
If it is a bijection then the inverse would be continuous too and range of the inverse, which is $(0,10)$ would be compact because continuous image of compact sets is compact, This is a contradiction. [A continuous bijection is strictly monotonic and this makes the inverse also monotonic and continuous].
answered Dec 7 '18 at 8:50
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
If it is a bijection then the inverse would be continuous too and range of the inverse, which is $(0,10)$ would be compact because continuous image of compact sets is compact, This is a contradiction. [A continuous bijection is strictly monotonic and this makes the inverse also monotonic and continuous].
answered Dec 7 '18 at 8:50
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
answered Dec 7 '18 at 8:50
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
answered Dec 7 '18 at 8:50
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
answered Dec 7 '18 at 8:50
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
53.5k32055
$begingroup$
@WilliamSun Continuous bijections on the real line are known to have continuous inverses automatically. I am assuming this result with a hint that this can b e proved using monotonicity.
$endgroup$
– Kavi Rama Murthy
Dec 7 '18 at 8:55
$begingroup$
It was very simple to under.stand. Thanks.
$endgroup$
– Mathsaddict
Dec 7 '18 at 9:04
add a comment |
$begingroup$
@WilliamSun Continuous bijections on the real line are known to have continuous inverses automatically. I am assuming this result with a hint that this can b e proved using monotonicity.
$endgroup$
– Kavi Rama Murthy
Dec 7 '18 at 8:55
$begingroup$
It was very simple to under.stand. Thanks.
$endgroup$
– Mathsaddict
Dec 7 '18 at 9:04
$begingroup$
@WilliamSun Continuous bijections on the real line are known to have continuous inverses automatically. I am assuming this result with a hint that this can b e proved using monotonicity.
$endgroup$
– Kavi Rama Murthy
Dec 7 '18 at 8:55
$begingroup$
@WilliamSun Continuous bijections on the real line are known to have continuous inverses automatically. I am assuming this result with a hint that this can b e proved using monotonicity.
$endgroup$
– Kavi Rama Murthy
Dec 7 '18 at 8:55
$begingroup$
It was very simple to under.stand. Thanks.
$endgroup$
– Mathsaddict
Dec 7 '18 at 9:04
$begingroup$
It was very simple to under.stand. Thanks.
$endgroup$
– Mathsaddict
Dec 7 '18 at 9:04
add a comment |
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