Mixing
How does the given proof is a spoof? [Completeness is a topological property of metric spaces]
Let $f:(X,d)to (Y,e)$ homeomorphism. So, $f^{-1}$ is continuous. So $f^{-1}$ is continuous at $y in Y$, For any $epsilon in mathbb R^+$, $exists deltain mathbb R^+$:$e(x,y)<delta implies d(f^{-1}(x),f^{-1}(y))<epsilon. $ Let ${y_n}$ be a cauchy sequence in $(Y,e)$. For the same $delta$, There is a $Nin mathbb N$: $forall m,nge Nimplies e(y_n,y_m)<delta$. Define a sequence ${x_n=f^{-1}(y_n)}$. $forall m,nge N implies d(f^{-1}(y_n),f^{-1}(y_m))<epsilon.$ Hence ${x_n}$ converges in $(X,d)$. So $f(x_n)$ converges to $f(x)$. Since $f$ is continuous. Hence ${y_n}$ converges to $f(x)$. How do my argument go wrong?
general-topology metric-spaces
asked Nov 20 '18 at 14:55
Math geek
37419
add a comment |
Let $f:(X,d)to (Y,e)$ homeomorphism. So, $f^{-1}$ is continuous. So $f^{-1}$ is continuous at $y in Y$, For any $epsilon in mathbb R^+$, $exists deltain mathbb R^+$:$e(x,y)<delta implies d(f^{-1}(x),f^{-1}(y))<epsilon. $ Let ${y_n}$ be a cauchy sequence in $(Y,e)$. For the same $delta$, There is a $Nin mathbb N$: $forall m,nge Nimplies e(y_n,y_m)<delta$. Define a sequence ${x_n=f^{-1}(y_n)}$. $forall m,nge N implies d(f^{-1}(y_n),f^{-1}(y_m))<epsilon.$ Hence ${x_n}$ converges in $(X,d)$. So $f(x_n)$ converges to $f(x)$. Since $f$ is continuous. Hence ${y_n}$ converges to $f(x)$. How do my argument go wrong?
general-topology metric-spaces
asked Nov 20 '18 at 14:55
Math geek
37419
Given $epsilon$, the $delta$ that you introduced will, in general, depend on $y$. Your argument needs the same $delta$ to work for infinitely many $y_n$'s.
– Andreas Blass
Nov 20 '18 at 15:26
add a comment |
Let $f:(X,d)to (Y,e)$ homeomorphism. So, $f^{-1}$ is continuous. So $f^{-1}$ is continuous at $y in Y$, For any $epsilon in mathbb R^+$, $exists deltain mathbb R^+$:$e(x,y)<delta implies d(f^{-1}(x),f^{-1}(y))<epsilon. $ Let ${y_n}$ be a cauchy sequence in $(Y,e)$. For the same $delta$, There is a $Nin mathbb N$: $forall m,nge Nimplies e(y_n,y_m)<delta$. Define a sequence ${x_n=f^{-1}(y_n)}$. $forall m,nge N implies d(f^{-1}(y_n),f^{-1}(y_m))<epsilon.$ Hence ${x_n}$ converges in $(X,d)$. So $f(x_n)$ converges to $f(x)$. Since $f$ is continuous. Hence ${y_n}$ converges to $f(x)$. How do my argument go wrong?
general-topology metric-spaces
asked Nov 20 '18 at 14:55
Math geek
37419
Let $f:(X,d)to (Y,e)$ homeomorphism. So, $f^{-1}$ is continuous. So $f^{-1}$ is continuous at $y in Y$, For any $epsilon in mathbb R^+$, $exists deltain mathbb R^+$:$e(x,y)<delta implies d(f^{-1}(x),f^{-1}(y))<epsilon. $ Let ${y_n}$ be a cauchy sequence in $(Y,e)$. For the same $delta$, There is a $Nin mathbb N$: $forall m,nge Nimplies e(y_n,y_m)<delta$. Define a sequence ${x_n=f^{-1}(y_n)}$. $forall m,nge N implies d(f^{-1}(y_n),f^{-1}(y_m))<epsilon.$ Hence ${x_n}$ converges in $(X,d)$. So $f(x_n)$ converges to $f(x)$. Since $f$ is continuous. Hence ${y_n}$ converges to $f(x)$. How do my argument go wrong?
general-topology metric-spaces
general-topology metric-spaces
asked Nov 20 '18 at 14:55
Math geek
37419
asked Nov 20 '18 at 14:55
Math geek
37419
asked Nov 20 '18 at 14:55
Math geek
37419
asked Nov 20 '18 at 14:55
Math geek
37419
asked Nov 20 '18 at 14:55
Math geek
37419
37419
Given $epsilon$, the $delta$ that you introduced will, in general, depend on $y$. Your argument needs the same $delta$ to work for infinitely many $y_n$'s.
– Andreas Blass
Nov 20 '18 at 15:26
add a comment |
Given $epsilon$, the $delta$ that you introduced will, in general, depend on $y$. Your argument needs the same $delta$ to work for infinitely many $y_n$'s.
– Andreas Blass
Nov 20 '18 at 15:26
Given $epsilon$, the $delta$ that you introduced will, in general, depend on $y$. Your argument needs the same $delta$ to work for infinitely many $y_n$'s.
– Andreas Blass
Nov 20 '18 at 15:26
Given $epsilon$, the $delta$ that you introduced will, in general, depend on $y$. Your argument needs the same $delta$ to work for infinitely many $y_n$'s.
– Andreas Blass
Nov 20 '18 at 15:26
add a comment |
1 Answer
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In your argument, you are using the uniform continuity criteria. In general, given $epsilon gt 0$, same $delta $ will not work for every point.
answered Nov 20 '18 at 15:13
Thomas Shelby
1,516216
add a comment |
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In your argument, you are using the uniform continuity criteria. In general, given $epsilon gt 0$, same $delta $ will not work for every point.
answered Nov 20 '18 at 15:13
Thomas Shelby
1,516216
add a comment |
In your argument, you are using the uniform continuity criteria. In general, given $epsilon gt 0$, same $delta $ will not work for every point.
answered Nov 20 '18 at 15:13
Thomas Shelby
1,516216
add a comment |
In your argument, you are using the uniform continuity criteria. In general, given $epsilon gt 0$, same $delta $ will not work for every point.
answered Nov 20 '18 at 15:13
Thomas Shelby
1,516216
In your argument, you are using the uniform continuity criteria. In general, given $epsilon gt 0$, same $delta $ will not work for every point.
answered Nov 20 '18 at 15:13
Thomas Shelby
1,516216
answered Nov 20 '18 at 15:13
Thomas Shelby
1,516216
answered Nov 20 '18 at 15:13
Thomas Shelby
1,516216
answered Nov 20 '18 at 15:13
Thomas Shelby
1,516216
1,516216
add a comment |
add a comment |
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Given $epsilon$, the $delta$ that you introduced will, in general, depend on $y$. Your argument needs the same $delta$ to work for infinitely many $y_n$'s.
– Andreas Blass
Nov 20 '18 at 15:26