Mixing
If $f:M mapsto M'$ is bijective and $d'(f(x),f(y))ge d(x,y)~forall x,yin M$ then $M$ compact $implies M'$...
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Let $(M,d)$ and $(M',d')$ be metric spaces and let $f:Mmapsto M'$ be a bijective function such that $$d'(f(x),f(y))ge d(x,y)~forall x,yin M$$
Is it true that if $M$ is compact then so is $M'$?
I found that the function $f(x)=begin{cases} x & text{ if } xin[0,1/2)\ x+1/2 & text{ if } xin[1/2,1]end{cases}$ satisfies the conditions for $M=[0,1]$ and $M'=[0,1/2)cup[1,3/2]$ and $M'$ is not compact.
Is this true? What if $f$ had to be continuous?
metric-spaces
asked Jan 30 at 12:45
John CataldoJohn Cataldo
1,1931316
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add a comment |
$begingroup$
Let $(M,d)$ and $(M',d')$ be metric spaces and let $f:Mmapsto M'$ be a bijective function such that $$d'(f(x),f(y))ge d(x,y)~forall x,yin M$$
Is it true that if $M$ is compact then so is $M'$?
I found that the function $f(x)=begin{cases} x & text{ if } xin[0,1/2)\ x+1/2 & text{ if } xin[1/2,1]end{cases}$ satisfies the conditions for $M=[0,1]$ and $M'=[0,1/2)cup[1,3/2]$ and $M'$ is not compact.
Is this true? What if $f$ had to be continuous?
metric-spaces
asked Jan 30 at 12:45
John CataldoJohn Cataldo
1,1931316
$endgroup$
add a comment |
$begingroup$
Let $(M,d)$ and $(M',d')$ be metric spaces and let $f:Mmapsto M'$ be a bijective function such that $$d'(f(x),f(y))ge d(x,y)~forall x,yin M$$
Is it true that if $M$ is compact then so is $M'$?
I found that the function $f(x)=begin{cases} x & text{ if } xin[0,1/2)\ x+1/2 & text{ if } xin[1/2,1]end{cases}$ satisfies the conditions for $M=[0,1]$ and $M'=[0,1/2)cup[1,3/2]$ and $M'$ is not compact.
Is this true? What if $f$ had to be continuous?
metric-spaces
asked Jan 30 at 12:45
John CataldoJohn Cataldo
1,1931316
$endgroup$
Let $(M,d)$ and $(M',d')$ be metric spaces and let $f:Mmapsto M'$ be a bijective function such that $$d'(f(x),f(y))ge d(x,y)~forall x,yin M$$
Is it true that if $M$ is compact then so is $M'$?
I found that the function $f(x)=begin{cases} x & text{ if } xin[0,1/2)\ x+1/2 & text{ if } xin[1/2,1]end{cases}$ satisfies the conditions for $M=[0,1]$ and $M'=[0,1/2)cup[1,3/2]$ and $M'$ is not compact.
Is this true? What if $f$ had to be continuous?
metric-spaces
metric-spaces
asked Jan 30 at 12:45
John CataldoJohn Cataldo
1,1931316
asked Jan 30 at 12:45
John CataldoJohn Cataldo
1,1931316
asked Jan 30 at 12:45
John CataldoJohn Cataldo
1,1931316
asked Jan 30 at 12:45
John CataldoJohn Cataldo
1,1931316
asked Jan 30 at 12:45
John CataldoJohn Cataldo
1,1931316
1,1931316
add a comment |
add a comment |
1 Answer
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Yes, your example works. If you're worried, then try going into more detail. How do you know $M$ is compact? (Cite a theorem.) How do you know $M'$ is not compact? (Find a sequence which contains no convergent subsequence, or indeed a non-convergent Cauchy sequence would do!) Are you sure that $d(x, y) le d'(f(x), f(y))$? Prove it!
If $f$ is continuous, then $M'$ must be compact, as the continuous image of a compact set is compact (regardless of whether $d(x, y) le d'(f(x), f(y))$ holds true or not).
answered Jan 30 at 12:51
Theo BenditTheo Bendit
20.2k12353
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Yes, your example works. If you're worried, then try going into more detail. How do you know $M$ is compact? (Cite a theorem.) How do you know $M'$ is not compact? (Find a sequence which contains no convergent subsequence, or indeed a non-convergent Cauchy sequence would do!) Are you sure that $d(x, y) le d'(f(x), f(y))$? Prove it!
If $f$ is continuous, then $M'$ must be compact, as the continuous image of a compact set is compact (regardless of whether $d(x, y) le d'(f(x), f(y))$ holds true or not).
answered Jan 30 at 12:51
Theo BenditTheo Bendit
20.2k12353
$endgroup$
add a comment |
$begingroup$
Yes, your example works. If you're worried, then try going into more detail. How do you know $M$ is compact? (Cite a theorem.) How do you know $M'$ is not compact? (Find a sequence which contains no convergent subsequence, or indeed a non-convergent Cauchy sequence would do!) Are you sure that $d(x, y) le d'(f(x), f(y))$? Prove it!
If $f$ is continuous, then $M'$ must be compact, as the continuous image of a compact set is compact (regardless of whether $d(x, y) le d'(f(x), f(y))$ holds true or not).
answered Jan 30 at 12:51
Theo BenditTheo Bendit
20.2k12353
$endgroup$
add a comment |
$begingroup$
Yes, your example works. If you're worried, then try going into more detail. How do you know $M$ is compact? (Cite a theorem.) How do you know $M'$ is not compact? (Find a sequence which contains no convergent subsequence, or indeed a non-convergent Cauchy sequence would do!) Are you sure that $d(x, y) le d'(f(x), f(y))$? Prove it!
If $f$ is continuous, then $M'$ must be compact, as the continuous image of a compact set is compact (regardless of whether $d(x, y) le d'(f(x), f(y))$ holds true or not).
answered Jan 30 at 12:51
Theo BenditTheo Bendit
20.2k12353
$endgroup$
Yes, your example works. If you're worried, then try going into more detail. How do you know $M$ is compact? (Cite a theorem.) How do you know $M'$ is not compact? (Find a sequence which contains no convergent subsequence, or indeed a non-convergent Cauchy sequence would do!) Are you sure that $d(x, y) le d'(f(x), f(y))$? Prove it!
If $f$ is continuous, then $M'$ must be compact, as the continuous image of a compact set is compact (regardless of whether $d(x, y) le d'(f(x), f(y))$ holds true or not).
answered Jan 30 at 12:51
Theo BenditTheo Bendit
20.2k12353
answered Jan 30 at 12:51
Theo BenditTheo Bendit
20.2k12353
answered Jan 30 at 12:51
Theo BenditTheo Bendit
20.2k12353
answered Jan 30 at 12:51
Theo BenditTheo Bendit
20.2k12353
20.2k12353
add a comment |
add a comment |
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