Mixing
Prove that an open interval (0,1) and a closed interval [0,1] are not homeomorphic.
$begingroup$
Prove that an open interval (0.1) and a closed interval [0,1] are not homeomorphic.
I am trying to prove this statement but only vague ideas on how to start.
Not using connectedness properties.
Please help
general-topology geometric-topology
asked Jul 25 '18 at 7:20
Nursultan NazariNursultan Nazari
111
$endgroup$
add a comment |
$begingroup$
Prove that an open interval (0.1) and a closed interval [0,1] are not homeomorphic.
I am trying to prove this statement but only vague ideas on how to start.
Not using connectedness properties.
Please help
general-topology geometric-topology
asked Jul 25 '18 at 7:20
Nursultan NazariNursultan Nazari
111
$endgroup$
1
$begingroup$
See this. Its really interesting!
$endgroup$
– user444830
Jul 25 '18 at 8:06
add a comment |
$begingroup$
Prove that an open interval (0.1) and a closed interval [0,1] are not homeomorphic.
I am trying to prove this statement but only vague ideas on how to start.
Not using connectedness properties.
Please help
general-topology geometric-topology
asked Jul 25 '18 at 7:20
Nursultan NazariNursultan Nazari
111
$endgroup$
Prove that an open interval (0.1) and a closed interval [0,1] are not homeomorphic.
I am trying to prove this statement but only vague ideas on how to start.
Not using connectedness properties.
Please help
general-topology geometric-topology
general-topology geometric-topology
asked Jul 25 '18 at 7:20
Nursultan NazariNursultan Nazari
111
asked Jul 25 '18 at 7:20
Nursultan NazariNursultan Nazari
111
asked Jul 25 '18 at 7:20
Nursultan NazariNursultan Nazari
111
asked Jul 25 '18 at 7:20
Nursultan NazariNursultan Nazari
111
asked Jul 25 '18 at 7:20
Nursultan NazariNursultan Nazari
111
111
1
$begingroup$
See this. Its really interesting!
$endgroup$
– user444830
Jul 25 '18 at 8:06
add a comment |
1
$begingroup$
See this. Its really interesting!
$endgroup$
– user444830
Jul 25 '18 at 8:06
1
1
$begingroup$
See this. Its really interesting!
$endgroup$
– user444830
Jul 25 '18 at 8:06
$begingroup$
See this. Its really interesting!
$endgroup$
– user444830
Jul 25 '18 at 8:06
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Consider the sequence $left(frac1nright)_{ninmathbb N}$ in $(0,1)$. It has no subsequence which converges to an element of $(0,1)$. However, every sequence of elements of $[0,1]$ has a subsequence that converges to an element of $[0,1]$, by the Bolzano-Weierstrass theorem and because $[0,1]$ is closed. Therefore, $(0,1)$ and $[0,1]$ are not homeomorphic.
answered Jul 25 '18 at 8:02
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
The open interval is not compact metric space, but the closed is.
answered Jul 25 '18 at 7:29
dmtridmtri
1,6612521
$endgroup$
$begingroup$
Thanks but it is not enough please show it
$endgroup$
– Nursultan Nazari
Jul 25 '18 at 7:35
$begingroup$
The continuous image of compact like the closed interval is compact... And the open interval is not compcat
$endgroup$
– dmtri
Jul 25 '18 at 7:41
add a comment |
$begingroup$
Suppose $f:[0,1]to(0,1)$ is a homeomorphism. In particular, then, it is a continuous surjection. Since $f$ is a continuous real-valued function on the closed interval $[0,1],$ it has an absolute minimum value, call it $m$. Then $min(0,1),$ and $frac m2in(0,1).$ Since $f$ is surjective, there is some $xin[0,1]$ such that $f(x)=frac m2lt m.$ This contradicts the fact that $m$ is the minimum value of $f.$
answered Jul 25 '18 at 7:36
bofbof
52.5k558121
$endgroup$
1
$begingroup$
Please correct the closed interval in your map
$endgroup$
– dmtri
Jul 25 '18 at 8:00
$begingroup$
@dmtri Oops. Thanks for the correction.
$endgroup$
– bof
Jul 25 '18 at 9:32
add a comment |
$begingroup$
Removing an endpoint (i.e. $0$ or $1$) from the closed interval $left[ 0,1right]$ leaves either $left( 0,1right]$ or $left[ 0,1right)$, both of which are connected spaces. On the other hand, removing any point from $(0,1)$ will leave a disconnected space (with two components). See the answer to this question.
This formalizes the feeling that $0$ and $1$ are "special points" in the space $left[0,1right]$. This kind of argument is often useful when dealing with simple spaces. For another example, see my answer to this question.
answered Jan 24 at 14:22
o.h.o.h.
6217
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the sequence $left(frac1nright)_{ninmathbb N}$ in $(0,1)$. It has no subsequence which converges to an element of $(0,1)$. However, every sequence of elements of $[0,1]$ has a subsequence that converges to an element of $[0,1]$, by the Bolzano-Weierstrass theorem and because $[0,1]$ is closed. Therefore, $(0,1)$ and $[0,1]$ are not homeomorphic.
answered Jul 25 '18 at 8:02
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
Consider the sequence $left(frac1nright)_{ninmathbb N}$ in $(0,1)$. It has no subsequence which converges to an element of $(0,1)$. However, every sequence of elements of $[0,1]$ has a subsequence that converges to an element of $[0,1]$, by the Bolzano-Weierstrass theorem and because $[0,1]$ is closed. Therefore, $(0,1)$ and $[0,1]$ are not homeomorphic.
answered Jul 25 '18 at 8:02
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
Consider the sequence $left(frac1nright)_{ninmathbb N}$ in $(0,1)$. It has no subsequence which converges to an element of $(0,1)$. However, every sequence of elements of $[0,1]$ has a subsequence that converges to an element of $[0,1]$, by the Bolzano-Weierstrass theorem and because $[0,1]$ is closed. Therefore, $(0,1)$ and $[0,1]$ are not homeomorphic.
answered Jul 25 '18 at 8:02
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
Consider the sequence $left(frac1nright)_{ninmathbb N}$ in $(0,1)$. It has no subsequence which converges to an element of $(0,1)$. However, every sequence of elements of $[0,1]$ has a subsequence that converges to an element of $[0,1]$, by the Bolzano-Weierstrass theorem and because $[0,1]$ is closed. Therefore, $(0,1)$ and $[0,1]$ are not homeomorphic.
answered Jul 25 '18 at 8:02
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jul 25 '18 at 8:02
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jul 25 '18 at 8:02
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jul 25 '18 at 8:02
José Carlos SantosJosé Carlos Santos
168k22132236
168k22132236
add a comment |
add a comment |
$begingroup$
The open interval is not compact metric space, but the closed is.
answered Jul 25 '18 at 7:29
dmtridmtri
1,6612521
$endgroup$
$begingroup$
Thanks but it is not enough please show it
$endgroup$
– Nursultan Nazari
Jul 25 '18 at 7:35
$begingroup$
The continuous image of compact like the closed interval is compact... And the open interval is not compcat
$endgroup$
– dmtri
Jul 25 '18 at 7:41
add a comment |
$begingroup$
The open interval is not compact metric space, but the closed is.
answered Jul 25 '18 at 7:29
dmtridmtri
1,6612521
$endgroup$
$begingroup$
Thanks but it is not enough please show it
$endgroup$
– Nursultan Nazari
Jul 25 '18 at 7:35
$begingroup$
The continuous image of compact like the closed interval is compact... And the open interval is not compcat
$endgroup$
– dmtri
Jul 25 '18 at 7:41
add a comment |
$begingroup$
The open interval is not compact metric space, but the closed is.
answered Jul 25 '18 at 7:29
dmtridmtri
1,6612521
$endgroup$
The open interval is not compact metric space, but the closed is.
answered Jul 25 '18 at 7:29
dmtridmtri
1,6612521
answered Jul 25 '18 at 7:29
dmtridmtri
1,6612521
answered Jul 25 '18 at 7:29
dmtridmtri
1,6612521
answered Jul 25 '18 at 7:29
dmtridmtri
1,6612521
1,6612521
$begingroup$
Thanks but it is not enough please show it
$endgroup$
– Nursultan Nazari
Jul 25 '18 at 7:35
$begingroup$
The continuous image of compact like the closed interval is compact... And the open interval is not compcat
$endgroup$
– dmtri
Jul 25 '18 at 7:41
add a comment |
$begingroup$
Thanks but it is not enough please show it
$endgroup$
– Nursultan Nazari
Jul 25 '18 at 7:35
$begingroup$
The continuous image of compact like the closed interval is compact... And the open interval is not compcat
$endgroup$
– dmtri
Jul 25 '18 at 7:41
$begingroup$
Thanks but it is not enough please show it
$endgroup$
– Nursultan Nazari
Jul 25 '18 at 7:35
$begingroup$
Thanks but it is not enough please show it
$endgroup$
– Nursultan Nazari
Jul 25 '18 at 7:35
$begingroup$
The continuous image of compact like the closed interval is compact... And the open interval is not compcat
$endgroup$
– dmtri
Jul 25 '18 at 7:41
$begingroup$
The continuous image of compact like the closed interval is compact... And the open interval is not compcat
$endgroup$
– dmtri
Jul 25 '18 at 7:41
add a comment |
$begingroup$
Suppose $f:[0,1]to(0,1)$ is a homeomorphism. In particular, then, it is a continuous surjection. Since $f$ is a continuous real-valued function on the closed interval $[0,1],$ it has an absolute minimum value, call it $m$. Then $min(0,1),$ and $frac m2in(0,1).$ Since $f$ is surjective, there is some $xin[0,1]$ such that $f(x)=frac m2lt m.$ This contradicts the fact that $m$ is the minimum value of $f.$
answered Jul 25 '18 at 7:36
bofbof
52.5k558121
$endgroup$
1
$begingroup$
Please correct the closed interval in your map
$endgroup$
– dmtri
Jul 25 '18 at 8:00
$begingroup$
@dmtri Oops. Thanks for the correction.
$endgroup$
– bof
Jul 25 '18 at 9:32
add a comment |
$begingroup$
Suppose $f:[0,1]to(0,1)$ is a homeomorphism. In particular, then, it is a continuous surjection. Since $f$ is a continuous real-valued function on the closed interval $[0,1],$ it has an absolute minimum value, call it $m$. Then $min(0,1),$ and $frac m2in(0,1).$ Since $f$ is surjective, there is some $xin[0,1]$ such that $f(x)=frac m2lt m.$ This contradicts the fact that $m$ is the minimum value of $f.$
answered Jul 25 '18 at 7:36
bofbof
52.5k558121
$endgroup$
1
$begingroup$
Please correct the closed interval in your map
$endgroup$
– dmtri
Jul 25 '18 at 8:00
$begingroup$
@dmtri Oops. Thanks for the correction.
$endgroup$
– bof
Jul 25 '18 at 9:32
add a comment |
$begingroup$
Suppose $f:[0,1]to(0,1)$ is a homeomorphism. In particular, then, it is a continuous surjection. Since $f$ is a continuous real-valued function on the closed interval $[0,1],$ it has an absolute minimum value, call it $m$. Then $min(0,1),$ and $frac m2in(0,1).$ Since $f$ is surjective, there is some $xin[0,1]$ such that $f(x)=frac m2lt m.$ This contradicts the fact that $m$ is the minimum value of $f.$
answered Jul 25 '18 at 7:36
bofbof
52.5k558121
$endgroup$
Suppose $f:[0,1]to(0,1)$ is a homeomorphism. In particular, then, it is a continuous surjection. Since $f$ is a continuous real-valued function on the closed interval $[0,1],$ it has an absolute minimum value, call it $m$. Then $min(0,1),$ and $frac m2in(0,1).$ Since $f$ is surjective, there is some $xin[0,1]$ such that $f(x)=frac m2lt m.$ This contradicts the fact that $m$ is the minimum value of $f.$
answered Jul 25 '18 at 7:36
bofbof
52.5k558121
edited Jul 25 '18 at 9:32
answered Jul 25 '18 at 7:36
bofbof
52.5k558121
answered Jul 25 '18 at 7:36
bofbof
52.5k558121
answered Jul 25 '18 at 7:36
bofbof
52.5k558121
52.5k558121
1
$begingroup$
Please correct the closed interval in your map
$endgroup$
– dmtri
Jul 25 '18 at 8:00
$begingroup$
@dmtri Oops. Thanks for the correction.
$endgroup$
– bof
Jul 25 '18 at 9:32
add a comment |
1
$begingroup$
Please correct the closed interval in your map
$endgroup$
– dmtri
Jul 25 '18 at 8:00
$begingroup$
@dmtri Oops. Thanks for the correction.
$endgroup$
– bof
Jul 25 '18 at 9:32
1
1
$begingroup$
Please correct the closed interval in your map
$endgroup$
– dmtri
Jul 25 '18 at 8:00
$begingroup$
Please correct the closed interval in your map
$endgroup$
– dmtri
Jul 25 '18 at 8:00
$begingroup$
@dmtri Oops. Thanks for the correction.
$endgroup$
– bof
Jul 25 '18 at 9:32
$begingroup$
@dmtri Oops. Thanks for the correction.
$endgroup$
– bof
Jul 25 '18 at 9:32
add a comment |
$begingroup$
Removing an endpoint (i.e. $0$ or $1$) from the closed interval $left[ 0,1right]$ leaves either $left( 0,1right]$ or $left[ 0,1right)$, both of which are connected spaces. On the other hand, removing any point from $(0,1)$ will leave a disconnected space (with two components). See the answer to this question.
This formalizes the feeling that $0$ and $1$ are "special points" in the space $left[0,1right]$. This kind of argument is often useful when dealing with simple spaces. For another example, see my answer to this question.
answered Jan 24 at 14:22
o.h.o.h.
6217
$endgroup$
add a comment |
$begingroup$
Removing an endpoint (i.e. $0$ or $1$) from the closed interval $left[ 0,1right]$ leaves either $left( 0,1right]$ or $left[ 0,1right)$, both of which are connected spaces. On the other hand, removing any point from $(0,1)$ will leave a disconnected space (with two components). See the answer to this question.
This formalizes the feeling that $0$ and $1$ are "special points" in the space $left[0,1right]$. This kind of argument is often useful when dealing with simple spaces. For another example, see my answer to this question.
answered Jan 24 at 14:22
o.h.o.h.
6217
$endgroup$
add a comment |
$begingroup$
Removing an endpoint (i.e. $0$ or $1$) from the closed interval $left[ 0,1right]$ leaves either $left( 0,1right]$ or $left[ 0,1right)$, both of which are connected spaces. On the other hand, removing any point from $(0,1)$ will leave a disconnected space (with two components). See the answer to this question.
This formalizes the feeling that $0$ and $1$ are "special points" in the space $left[0,1right]$. This kind of argument is often useful when dealing with simple spaces. For another example, see my answer to this question.
answered Jan 24 at 14:22
o.h.o.h.
6217
$endgroup$
Removing an endpoint (i.e. $0$ or $1$) from the closed interval $left[ 0,1right]$ leaves either $left( 0,1right]$ or $left[ 0,1right)$, both of which are connected spaces. On the other hand, removing any point from $(0,1)$ will leave a disconnected space (with two components). See the answer to this question.
This formalizes the feeling that $0$ and $1$ are "special points" in the space $left[0,1right]$. This kind of argument is often useful when dealing with simple spaces. For another example, see my answer to this question.
answered Jan 24 at 14:22
o.h.o.h.
6217
answered Jan 24 at 14:22
o.h.o.h.
6217
answered Jan 24 at 14:22
o.h.o.h.
6217
answered Jan 24 at 14:22
o.h.o.h.
6217
6217
add a comment |
add a comment |
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$begingroup$
See this. Its really interesting!
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– user444830
Jul 25 '18 at 8:06