Intermediate Value Theorem in point-set topology.
$begingroup$
Theorem: Let $f:X to Y$ be a continuous map, where $X$ is a connected space and Y is an ordered set in the order topology. If $a,b in X$ and $r in Y$ such that $f(a) lt r lt f(b)$, then $exists c in X (f(c)=r)$
Proof Attempt: Let $f:X to Y$ be a continuous map where $X$ is a connected space and $Y$ is an ordered set under the order topology. Suppose $a,b in X$ and $r in Y$ such that $f(a) lt r lt f(b)$. Since $X$ is connected then $f(X)$ is connected. Let $f(X)=(gamma_0, gamma)cup (overline{gamma},gamma')$. Since $f(a),f(b) in f(X) subset Y$, then $f(a),f(b)in (gamma_0, gamma)cup (overline{gamma},gamma')$, then, w.l.o.g., let $f(b)in (overline{gamma},gamma')$, and $f(a)in (overline{gamma},gamma')$. Also, $f(X)$ is connected so $(gamma_0,gamma)cap (gamma,gamma')neqemptyset$. This forces $rin(gamma_0,gamma)cup (overline{gamma},gamma')=f(X)$. Therefore, $rin f(X)$ and $exists cin X(f(c)=r)$.
general-topology proof-verification
asked Jan 6 at 12:44
TheLast CipherTheLast Cipher
715715
$endgroup$
add a comment |
$begingroup$
Theorem: Let $f:X to Y$ be a continuous map, where $X$ is a connected space and Y is an ordered set in the order topology. If $a,b in X$ and $r in Y$ such that $f(a) lt r lt f(b)$, then $exists c in X (f(c)=r)$
Proof Attempt: Let $f:X to Y$ be a continuous map where $X$ is a connected space and $Y$ is an ordered set under the order topology. Suppose $a,b in X$ and $r in Y$ such that $f(a) lt r lt f(b)$. Since $X$ is connected then $f(X)$ is connected. Let $f(X)=(gamma_0, gamma)cup (overline{gamma},gamma')$. Since $f(a),f(b) in f(X) subset Y$, then $f(a),f(b)in (gamma_0, gamma)cup (overline{gamma},gamma')$, then, w.l.o.g., let $f(b)in (overline{gamma},gamma')$, and $f(a)in (overline{gamma},gamma')$. Also, $f(X)$ is connected so $(gamma_0,gamma)cap (gamma,gamma')neqemptyset$. This forces $rin(gamma_0,gamma)cup (overline{gamma},gamma')=f(X)$. Therefore, $rin f(X)$ and $exists cin X(f(c)=r)$.
general-topology proof-verification
asked Jan 6 at 12:44
TheLast CipherTheLast Cipher
715715
$endgroup$
$begingroup$
What is the question?
$endgroup$
– Wojowu
Jan 6 at 12:46
$begingroup$
I want the proof I wrote to be verified.
$endgroup$
– TheLast Cipher
Jan 6 at 12:46
$begingroup$
Could you at least let us know what you are proving? Please give the complete statement.
$endgroup$
– Wojowu
Jan 6 at 12:47
$begingroup$
I have added the theorem's statement. Thanks.
$endgroup$
– TheLast Cipher
Jan 6 at 12:52
$begingroup$
why is $f[X]$ of that form? I see no justification. Just a lot of useless $gamma$'s I think.
$endgroup$
– Henno Brandsma
Jan 6 at 13:27
add a comment |
$begingroup$
Theorem: Let $f:X to Y$ be a continuous map, where $X$ is a connected space and Y is an ordered set in the order topology. If $a,b in X$ and $r in Y$ such that $f(a) lt r lt f(b)$, then $exists c in X (f(c)=r)$
Proof Attempt: Let $f:X to Y$ be a continuous map where $X$ is a connected space and $Y$ is an ordered set under the order topology. Suppose $a,b in X$ and $r in Y$ such that $f(a) lt r lt f(b)$. Since $X$ is connected then $f(X)$ is connected. Let $f(X)=(gamma_0, gamma)cup (overline{gamma},gamma')$. Since $f(a),f(b) in f(X) subset Y$, then $f(a),f(b)in (gamma_0, gamma)cup (overline{gamma},gamma')$, then, w.l.o.g., let $f(b)in (overline{gamma},gamma')$, and $f(a)in (overline{gamma},gamma')$. Also, $f(X)$ is connected so $(gamma_0,gamma)cap (gamma,gamma')neqemptyset$. This forces $rin(gamma_0,gamma)cup (overline{gamma},gamma')=f(X)$. Therefore, $rin f(X)$ and $exists cin X(f(c)=r)$.
general-topology proof-verification
asked Jan 6 at 12:44
TheLast CipherTheLast Cipher
715715
$endgroup$
Theorem: Let $f:X to Y$ be a continuous map, where $X$ is a connected space and Y is an ordered set in the order topology. If $a,b in X$ and $r in Y$ such that $f(a) lt r lt f(b)$, then $exists c in X (f(c)=r)$
Proof Attempt: Let $f:X to Y$ be a continuous map where $X$ is a connected space and $Y$ is an ordered set under the order topology. Suppose $a,b in X$ and $r in Y$ such that $f(a) lt r lt f(b)$. Since $X$ is connected then $f(X)$ is connected. Let $f(X)=(gamma_0, gamma)cup (overline{gamma},gamma')$. Since $f(a),f(b) in f(X) subset Y$, then $f(a),f(b)in (gamma_0, gamma)cup (overline{gamma},gamma')$, then, w.l.o.g., let $f(b)in (overline{gamma},gamma')$, and $f(a)in (overline{gamma},gamma')$. Also, $f(X)$ is connected so $(gamma_0,gamma)cap (gamma,gamma')neqemptyset$. This forces $rin(gamma_0,gamma)cup (overline{gamma},gamma')=f(X)$. Therefore, $rin f(X)$ and $exists cin X(f(c)=r)$.
general-topology proof-verification
general-topology proof-verification
asked Jan 6 at 12:44
TheLast CipherTheLast Cipher
715715
asked Jan 6 at 12:44
TheLast CipherTheLast Cipher
715715
edited Jan 6 at 12:52
TheLast Cipher
asked Jan 6 at 12:44
TheLast CipherTheLast Cipher
715715
asked Jan 6 at 12:44
TheLast CipherTheLast Cipher
715715
asked Jan 6 at 12:44
TheLast CipherTheLast Cipher
715715
715715
$begingroup$
What is the question?
$endgroup$
– Wojowu
Jan 6 at 12:46
$begingroup$
I want the proof I wrote to be verified.
$endgroup$
– TheLast Cipher
Jan 6 at 12:46
$begingroup$
Could you at least let us know what you are proving? Please give the complete statement.
$endgroup$
– Wojowu
Jan 6 at 12:47
$begingroup$
I have added the theorem's statement. Thanks.
$endgroup$
– TheLast Cipher
Jan 6 at 12:52
$begingroup$
why is $f[X]$ of that form? I see no justification. Just a lot of useless $gamma$'s I think.
$endgroup$
– Henno Brandsma
Jan 6 at 13:27
add a comment |
$begingroup$
What is the question?
$endgroup$
– Wojowu
Jan 6 at 12:46
$begingroup$
I want the proof I wrote to be verified.
$endgroup$
– TheLast Cipher
Jan 6 at 12:46
$begingroup$
Could you at least let us know what you are proving? Please give the complete statement.
$endgroup$
– Wojowu
Jan 6 at 12:47
$begingroup$
I have added the theorem's statement. Thanks.
$endgroup$
– TheLast Cipher
Jan 6 at 12:52
$begingroup$
why is $f[X]$ of that form? I see no justification. Just a lot of useless $gamma$'s I think.
$endgroup$
– Henno Brandsma
Jan 6 at 13:27
$begingroup$
What is the question?
$endgroup$
– Wojowu
Jan 6 at 12:46
$begingroup$
What is the question?
$endgroup$
– Wojowu
Jan 6 at 12:46
$begingroup$
I want the proof I wrote to be verified.
$endgroup$
– TheLast Cipher
Jan 6 at 12:46
$begingroup$
I want the proof I wrote to be verified.
$endgroup$
– TheLast Cipher
Jan 6 at 12:46
$begingroup$
Could you at least let us know what you are proving? Please give the complete statement.
$endgroup$
– Wojowu
Jan 6 at 12:47
$begingroup$
Could you at least let us know what you are proving? Please give the complete statement.
$endgroup$
– Wojowu
Jan 6 at 12:47
$begingroup$
I have added the theorem's statement. Thanks.
$endgroup$
– TheLast Cipher
Jan 6 at 12:52
$begingroup$
I have added the theorem's statement. Thanks.
$endgroup$
– TheLast Cipher
Jan 6 at 12:52
$begingroup$
why is $f[X]$ of that form? I see no justification. Just a lot of useless $gamma$'s I think.
$endgroup$
– Henno Brandsma
Jan 6 at 13:27
$begingroup$
why is $f[X]$ of that form? I see no justification. Just a lot of useless $gamma$'s I think.
$endgroup$
– Henno Brandsma
Jan 6 at 13:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Better: suppose we have $(Y,<)$ in the order topology and $f: X to Y$ continuous with $X$ connected and for some points $a,b in X$ we have $f(a) < r < f(b)$ with $r in Y$.
Suppose there is no $x in X$ with $f(x)=r$.
Then define $O_1={y in Y: y < r }$ and $O_2={y in Y: y > r}$ which are open in $Y$ in the order topology.
Firstly $$f^{-1}[O_1] cup f^{-1}[O_2] = X$$
because we assumed $r$ is not assumed as a value, and $Y$ has a linear order, so always $f(x) > r$ or $f(x) < r$ must hold, but not both, so the sets are disjoint. We also know that $a in f^{-1}[O_1]$ and $b in f^{-1}[O_2]$, so both these sets are open (by continity of $f$), disjoint, non-empty and cover $X$. This is a contradiction with the connectedness of $X$. So we must have some $x$ with $f(x)=r$.
answered Jan 6 at 13:26
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
I initially thought of starting with the $f^{-1}[O_1]$ and $f^{-1}[O_2]$ like how you have defined it. But I think that would make $f(X)$ not connected, since $X$ is connected, would it not?
$endgroup$
– TheLast Cipher
Jan 6 at 14:08
$begingroup$
@TheLastCipher it doesn’t matter really if we get a contradiction from $X$ connected or $f[X]$ connected. The eventual idea will be the same.
$endgroup$
– Henno Brandsma
Jan 6 at 14:19
$begingroup$
I don't understand why it would not matter, especially when we use it as our starting assumption. :(
$endgroup$
– TheLast Cipher
Jan 6 at 14:23
$begingroup$
@TheLastCipher I chose to use the assumption "$X$ connected" directly. $O_1$ and $O_2$ form a direct disconnection of $f[X]$ which also gives a contradiction with a known theorem. Both are OK. My proof can be seen as a partial reproof of $X$ connected implies $f[X]$ connected in a way.
$endgroup$
– Henno Brandsma
Jan 6 at 14:27
$begingroup$
ahh yes. It's the contrapositive. Thanks!
$endgroup$
– TheLast Cipher
Jan 6 at 14:30
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
Better: suppose we have $(Y,<)$ in the order topology and $f: X to Y$ continuous with $X$ connected and for some points $a,b in X$ we have $f(a) < r < f(b)$ with $r in Y$.
Suppose there is no $x in X$ with $f(x)=r$.
Then define $O_1={y in Y: y < r }$ and $O_2={y in Y: y > r}$ which are open in $Y$ in the order topology.
Firstly $$f^{-1}[O_1] cup f^{-1}[O_2] = X$$
because we assumed $r$ is not assumed as a value, and $Y$ has a linear order, so always $f(x) > r$ or $f(x) < r$ must hold, but not both, so the sets are disjoint. We also know that $a in f^{-1}[O_1]$ and $b in f^{-1}[O_2]$, so both these sets are open (by continity of $f$), disjoint, non-empty and cover $X$. This is a contradiction with the connectedness of $X$. So we must have some $x$ with $f(x)=r$.
answered Jan 6 at 13:26
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
I initially thought of starting with the $f^{-1}[O_1]$ and $f^{-1}[O_2]$ like how you have defined it. But I think that would make $f(X)$ not connected, since $X$ is connected, would it not?
$endgroup$
– TheLast Cipher
Jan 6 at 14:08
$begingroup$
@TheLastCipher it doesn’t matter really if we get a contradiction from $X$ connected or $f[X]$ connected. The eventual idea will be the same.
$endgroup$
– Henno Brandsma
Jan 6 at 14:19
$begingroup$
I don't understand why it would not matter, especially when we use it as our starting assumption. :(
$endgroup$
– TheLast Cipher
Jan 6 at 14:23
$begingroup$
@TheLastCipher I chose to use the assumption "$X$ connected" directly. $O_1$ and $O_2$ form a direct disconnection of $f[X]$ which also gives a contradiction with a known theorem. Both are OK. My proof can be seen as a partial reproof of $X$ connected implies $f[X]$ connected in a way.
$endgroup$
– Henno Brandsma
Jan 6 at 14:27
$begingroup$
ahh yes. It's the contrapositive. Thanks!
$endgroup$
– TheLast Cipher
Jan 6 at 14:30
|
show 1 more comment
$begingroup$
Better: suppose we have $(Y,<)$ in the order topology and $f: X to Y$ continuous with $X$ connected and for some points $a,b in X$ we have $f(a) < r < f(b)$ with $r in Y$.
Suppose there is no $x in X$ with $f(x)=r$.
Then define $O_1={y in Y: y < r }$ and $O_2={y in Y: y > r}$ which are open in $Y$ in the order topology.
Firstly $$f^{-1}[O_1] cup f^{-1}[O_2] = X$$
because we assumed $r$ is not assumed as a value, and $Y$ has a linear order, so always $f(x) > r$ or $f(x) < r$ must hold, but not both, so the sets are disjoint. We also know that $a in f^{-1}[O_1]$ and $b in f^{-1}[O_2]$, so both these sets are open (by continity of $f$), disjoint, non-empty and cover $X$. This is a contradiction with the connectedness of $X$. So we must have some $x$ with $f(x)=r$.
answered Jan 6 at 13:26
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
I initially thought of starting with the $f^{-1}[O_1]$ and $f^{-1}[O_2]$ like how you have defined it. But I think that would make $f(X)$ not connected, since $X$ is connected, would it not?
$endgroup$
– TheLast Cipher
Jan 6 at 14:08
$begingroup$
@TheLastCipher it doesn’t matter really if we get a contradiction from $X$ connected or $f[X]$ connected. The eventual idea will be the same.
$endgroup$
– Henno Brandsma
Jan 6 at 14:19
$begingroup$
I don't understand why it would not matter, especially when we use it as our starting assumption. :(
$endgroup$
– TheLast Cipher
Jan 6 at 14:23
$begingroup$
@TheLastCipher I chose to use the assumption "$X$ connected" directly. $O_1$ and $O_2$ form a direct disconnection of $f[X]$ which also gives a contradiction with a known theorem. Both are OK. My proof can be seen as a partial reproof of $X$ connected implies $f[X]$ connected in a way.
$endgroup$
– Henno Brandsma
Jan 6 at 14:27
$begingroup$
ahh yes. It's the contrapositive. Thanks!
$endgroup$
– TheLast Cipher
Jan 6 at 14:30
|
show 1 more comment
$begingroup$
Better: suppose we have $(Y,<)$ in the order topology and $f: X to Y$ continuous with $X$ connected and for some points $a,b in X$ we have $f(a) < r < f(b)$ with $r in Y$.
Suppose there is no $x in X$ with $f(x)=r$.
Then define $O_1={y in Y: y < r }$ and $O_2={y in Y: y > r}$ which are open in $Y$ in the order topology.
Firstly $$f^{-1}[O_1] cup f^{-1}[O_2] = X$$
because we assumed $r$ is not assumed as a value, and $Y$ has a linear order, so always $f(x) > r$ or $f(x) < r$ must hold, but not both, so the sets are disjoint. We also know that $a in f^{-1}[O_1]$ and $b in f^{-1}[O_2]$, so both these sets are open (by continity of $f$), disjoint, non-empty and cover $X$. This is a contradiction with the connectedness of $X$. So we must have some $x$ with $f(x)=r$.
answered Jan 6 at 13:26
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
Better: suppose we have $(Y,<)$ in the order topology and $f: X to Y$ continuous with $X$ connected and for some points $a,b in X$ we have $f(a) < r < f(b)$ with $r in Y$.
Suppose there is no $x in X$ with $f(x)=r$.
Then define $O_1={y in Y: y < r }$ and $O_2={y in Y: y > r}$ which are open in $Y$ in the order topology.
Firstly $$f^{-1}[O_1] cup f^{-1}[O_2] = X$$
because we assumed $r$ is not assumed as a value, and $Y$ has a linear order, so always $f(x) > r$ or $f(x) < r$ must hold, but not both, so the sets are disjoint. We also know that $a in f^{-1}[O_1]$ and $b in f^{-1}[O_2]$, so both these sets are open (by continity of $f$), disjoint, non-empty and cover $X$. This is a contradiction with the connectedness of $X$. So we must have some $x$ with $f(x)=r$.
answered Jan 6 at 13:26
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 6 at 13:26
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 6 at 13:26
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 6 at 13:26
Henno BrandsmaHenno Brandsma
113k348123
113k348123
$begingroup$
I initially thought of starting with the $f^{-1}[O_1]$ and $f^{-1}[O_2]$ like how you have defined it. But I think that would make $f(X)$ not connected, since $X$ is connected, would it not?
$endgroup$
– TheLast Cipher
Jan 6 at 14:08
$begingroup$
@TheLastCipher it doesn’t matter really if we get a contradiction from $X$ connected or $f[X]$ connected. The eventual idea will be the same.
$endgroup$
– Henno Brandsma
Jan 6 at 14:19
$begingroup$
I don't understand why it would not matter, especially when we use it as our starting assumption. :(
$endgroup$
– TheLast Cipher
Jan 6 at 14:23
$begingroup$
@TheLastCipher I chose to use the assumption "$X$ connected" directly. $O_1$ and $O_2$ form a direct disconnection of $f[X]$ which also gives a contradiction with a known theorem. Both are OK. My proof can be seen as a partial reproof of $X$ connected implies $f[X]$ connected in a way.
$endgroup$
– Henno Brandsma
Jan 6 at 14:27
$begingroup$
ahh yes. It's the contrapositive. Thanks!
$endgroup$
– TheLast Cipher
Jan 6 at 14:30
|
show 1 more comment
$begingroup$
I initially thought of starting with the $f^{-1}[O_1]$ and $f^{-1}[O_2]$ like how you have defined it. But I think that would make $f(X)$ not connected, since $X$ is connected, would it not?
$endgroup$
– TheLast Cipher
Jan 6 at 14:08
$begingroup$
@TheLastCipher it doesn’t matter really if we get a contradiction from $X$ connected or $f[X]$ connected. The eventual idea will be the same.
$endgroup$
– Henno Brandsma
Jan 6 at 14:19
$begingroup$
I don't understand why it would not matter, especially when we use it as our starting assumption. :(
$endgroup$
– TheLast Cipher
Jan 6 at 14:23
$begingroup$
@TheLastCipher I chose to use the assumption "$X$ connected" directly. $O_1$ and $O_2$ form a direct disconnection of $f[X]$ which also gives a contradiction with a known theorem. Both are OK. My proof can be seen as a partial reproof of $X$ connected implies $f[X]$ connected in a way.
$endgroup$
– Henno Brandsma
Jan 6 at 14:27
$begingroup$
ahh yes. It's the contrapositive. Thanks!
$endgroup$
– TheLast Cipher
Jan 6 at 14:30
$begingroup$
I initially thought of starting with the $f^{-1}[O_1]$ and $f^{-1}[O_2]$ like how you have defined it. But I think that would make $f(X)$ not connected, since $X$ is connected, would it not?
$endgroup$
– TheLast Cipher
Jan 6 at 14:08
$begingroup$
I initially thought of starting with the $f^{-1}[O_1]$ and $f^{-1}[O_2]$ like how you have defined it. But I think that would make $f(X)$ not connected, since $X$ is connected, would it not?
$endgroup$
– TheLast Cipher
Jan 6 at 14:08
$begingroup$
@TheLastCipher it doesn’t matter really if we get a contradiction from $X$ connected or $f[X]$ connected. The eventual idea will be the same.
$endgroup$
– Henno Brandsma
Jan 6 at 14:19
$begingroup$
@TheLastCipher it doesn’t matter really if we get a contradiction from $X$ connected or $f[X]$ connected. The eventual idea will be the same.
$endgroup$
– Henno Brandsma
Jan 6 at 14:19
$begingroup$
I don't understand why it would not matter, especially when we use it as our starting assumption. :(
$endgroup$
– TheLast Cipher
Jan 6 at 14:23
$begingroup$
I don't understand why it would not matter, especially when we use it as our starting assumption. :(
$endgroup$
– TheLast Cipher
Jan 6 at 14:23
$begingroup$
@TheLastCipher I chose to use the assumption "$X$ connected" directly. $O_1$ and $O_2$ form a direct disconnection of $f[X]$ which also gives a contradiction with a known theorem. Both are OK. My proof can be seen as a partial reproof of $X$ connected implies $f[X]$ connected in a way.
$endgroup$
– Henno Brandsma
Jan 6 at 14:27
$begingroup$
@TheLastCipher I chose to use the assumption "$X$ connected" directly. $O_1$ and $O_2$ form a direct disconnection of $f[X]$ which also gives a contradiction with a known theorem. Both are OK. My proof can be seen as a partial reproof of $X$ connected implies $f[X]$ connected in a way.
$endgroup$
– Henno Brandsma
Jan 6 at 14:27
$begingroup$
ahh yes. It's the contrapositive. Thanks!
$endgroup$
– TheLast Cipher
Jan 6 at 14:30
$begingroup$
ahh yes. It's the contrapositive. Thanks!
$endgroup$
– TheLast Cipher
Jan 6 at 14:30
|
show 1 more comment
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$begingroup$
What is the question?
$endgroup$
– Wojowu
Jan 6 at 12:46
$begingroup$
I want the proof I wrote to be verified.
$endgroup$
– TheLast Cipher
Jan 6 at 12:46
$begingroup$
Could you at least let us know what you are proving? Please give the complete statement.
$endgroup$
– Wojowu
Jan 6 at 12:47
$begingroup$
I have added the theorem's statement. Thanks.
$endgroup$
– TheLast Cipher
Jan 6 at 12:52
$begingroup$
why is $f[X]$ of that form? I see no justification. Just a lot of useless $gamma$'s I think.
$endgroup$
– Henno Brandsma
Jan 6 at 13:27