Mixing
Does Intermediate value theorem work depending on the domain
$begingroup$
Does it work when the domain and codomain are subsets of the reals? For example, any continuous function from a discrete subset of reals that satisfies the least upper bound property? Is this possible? If it breaks down, why?
real-analysis functional-analysis continuity
asked Jan 20 at 15:10
Dis-integratingDis-integrating
1,038426
$endgroup$
add a comment |
$begingroup$
Does it work when the domain and codomain are subsets of the reals? For example, any continuous function from a discrete subset of reals that satisfies the least upper bound property? Is this possible? If it breaks down, why?
real-analysis functional-analysis continuity
asked Jan 20 at 15:10
Dis-integratingDis-integrating
1,038426
$endgroup$
$begingroup$
Because of trivial examples such as $Bbb Zto Bbb R$, $xmapsto x$
$endgroup$
– Hagen von Eitzen
Jan 20 at 15:14
add a comment |
$begingroup$
Does it work when the domain and codomain are subsets of the reals? For example, any continuous function from a discrete subset of reals that satisfies the least upper bound property? Is this possible? If it breaks down, why?
real-analysis functional-analysis continuity
asked Jan 20 at 15:10
Dis-integratingDis-integrating
1,038426
$endgroup$
Does it work when the domain and codomain are subsets of the reals? For example, any continuous function from a discrete subset of reals that satisfies the least upper bound property? Is this possible? If it breaks down, why?
real-analysis functional-analysis continuity
real-analysis functional-analysis continuity
asked Jan 20 at 15:10
Dis-integratingDis-integrating
1,038426
asked Jan 20 at 15:10
Dis-integratingDis-integrating
1,038426
asked Jan 20 at 15:10
Dis-integratingDis-integrating
1,038426
asked Jan 20 at 15:10
Dis-integratingDis-integrating
1,038426
asked Jan 20 at 15:10
Dis-integratingDis-integrating
1,038426
1,038426
$begingroup$
Because of trivial examples such as $Bbb Zto Bbb R$, $xmapsto x$
$endgroup$
– Hagen von Eitzen
Jan 20 at 15:14
add a comment |
$begingroup$
Because of trivial examples such as $Bbb Zto Bbb R$, $xmapsto x$
$endgroup$
– Hagen von Eitzen
Jan 20 at 15:14
$begingroup$
Because of trivial examples such as $Bbb Zto Bbb R$, $xmapsto x$
$endgroup$
– Hagen von Eitzen
Jan 20 at 15:14
$begingroup$
Because of trivial examples such as $Bbb Zto Bbb R$, $xmapsto x$
$endgroup$
– Hagen von Eitzen
Jan 20 at 15:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's say that a subset $Bbb X$ of $Bbb R$ has the IVT property if for every continuous map $fcolon Xto Bbb R$ and every $x_1,x_2in X$, $yin Bbb R$ with $f(x_1)<y<f(x_2)$, there exists $xiin X$ with $f(xi)=y$.
Then $X$ has the IVT property iff $X$ is connected.
Assume $X$ has the IVT property. Let $U,Vsubseteq Bbb R$ be open sets such that $Xsubseteq Ucup V$ and the sets $Xcap U$ and $Xcap V$ are disjoint. Define $fcolon XtoBbb R$ as $f(x)=begin{cases}1&xin U\0&text{otherwise}end{cases}$. Then $f$ is continuous. If there were $x_1,x_2in X$ with $f(x_1)=0$ and $f(x_2)=1$, then the IVT property would tell us that $f(xi)=frac12$ for some $xiin X$. As this is absurd, such $x_1,x_2$ cannot exist, i.e., one of $Ucap X$, $Vcap X$ is empty. We conclude that $X$ is connected.
Assume $X$ is connected. Let $fcolon Xto Bbb R$ be continuous, $x_1,x_2in X$, $yin Bbb R$ with $f(x_1)<y<f(x_2)$. Then the sets $f^{-1}((-infty,y))$ and $f^{-1}((y,infty))$ are disjoint and, by continuity of $f$, open subsets of $X$, They are also non.empty (contain $x_1$ and $x_2$, respectively). By connectedness of $X$ they thus cannot cover $X$. Hence there must exist $xiin X$ such that $f(xi)notin(-infty,y)cup(y,infty)$, i.e., $f(xi)=y$.
Quite obviously, discrete subsets of $Bbb R$ (with at least two points) are not connected. Of course you can also see immediately that for any such discrete (or "gappy" in any other way) subset of $Bbb R$, the continuous map $xmapsto x$ breaks the Intermediate value prperty.
answered Jan 20 at 15:30
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
$begingroup$
Why can we assume that there exist two open sets $U, V$ such that $Xcap U$ and $Xcap V$ are disjoint? thanks in advance.
$endgroup$
– StammeringMathematician
Jan 20 at 16:02
add a comment |
$begingroup$
No. It is not true that every function from $mathbb Z$ into $mathbb R$ satisfies the intermediate value property (note that every function from $mathbb Z$ into $mathbb R$ is continuous), in spite of the fact that $mathbb Z$ is discrete and that it satisfies the least upper bound property.
answered Jan 20 at 15:15
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
all subsets of Z don't satisfy l.u.b?
$endgroup$
– Dis-integrating
Jan 20 at 16:00
$begingroup$
Yes they do: a non-emtpy subset of $mathbb Z$ with an upper bound always has a least upper bound.
$endgroup$
– José Carlos Santos
Jan 20 at 16:04
$begingroup$
oh right yes, we need that it has an upper bound in the proof anyway
$endgroup$
– Dis-integrating
Jan 20 at 16:05
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
Let's say that a subset $Bbb X$ of $Bbb R$ has the IVT property if for every continuous map $fcolon Xto Bbb R$ and every $x_1,x_2in X$, $yin Bbb R$ with $f(x_1)<y<f(x_2)$, there exists $xiin X$ with $f(xi)=y$.
Then $X$ has the IVT property iff $X$ is connected.
Assume $X$ has the IVT property. Let $U,Vsubseteq Bbb R$ be open sets such that $Xsubseteq Ucup V$ and the sets $Xcap U$ and $Xcap V$ are disjoint. Define $fcolon XtoBbb R$ as $f(x)=begin{cases}1&xin U\0&text{otherwise}end{cases}$. Then $f$ is continuous. If there were $x_1,x_2in X$ with $f(x_1)=0$ and $f(x_2)=1$, then the IVT property would tell us that $f(xi)=frac12$ for some $xiin X$. As this is absurd, such $x_1,x_2$ cannot exist, i.e., one of $Ucap X$, $Vcap X$ is empty. We conclude that $X$ is connected.
Assume $X$ is connected. Let $fcolon Xto Bbb R$ be continuous, $x_1,x_2in X$, $yin Bbb R$ with $f(x_1)<y<f(x_2)$. Then the sets $f^{-1}((-infty,y))$ and $f^{-1}((y,infty))$ are disjoint and, by continuity of $f$, open subsets of $X$, They are also non.empty (contain $x_1$ and $x_2$, respectively). By connectedness of $X$ they thus cannot cover $X$. Hence there must exist $xiin X$ such that $f(xi)notin(-infty,y)cup(y,infty)$, i.e., $f(xi)=y$.
Quite obviously, discrete subsets of $Bbb R$ (with at least two points) are not connected. Of course you can also see immediately that for any such discrete (or "gappy" in any other way) subset of $Bbb R$, the continuous map $xmapsto x$ breaks the Intermediate value prperty.
answered Jan 20 at 15:30
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
$begingroup$
Why can we assume that there exist two open sets $U, V$ such that $Xcap U$ and $Xcap V$ are disjoint? thanks in advance.
$endgroup$
– StammeringMathematician
Jan 20 at 16:02
add a comment |
$begingroup$
Let's say that a subset $Bbb X$ of $Bbb R$ has the IVT property if for every continuous map $fcolon Xto Bbb R$ and every $x_1,x_2in X$, $yin Bbb R$ with $f(x_1)<y<f(x_2)$, there exists $xiin X$ with $f(xi)=y$.
Then $X$ has the IVT property iff $X$ is connected.
Assume $X$ has the IVT property. Let $U,Vsubseteq Bbb R$ be open sets such that $Xsubseteq Ucup V$ and the sets $Xcap U$ and $Xcap V$ are disjoint. Define $fcolon XtoBbb R$ as $f(x)=begin{cases}1&xin U\0&text{otherwise}end{cases}$. Then $f$ is continuous. If there were $x_1,x_2in X$ with $f(x_1)=0$ and $f(x_2)=1$, then the IVT property would tell us that $f(xi)=frac12$ for some $xiin X$. As this is absurd, such $x_1,x_2$ cannot exist, i.e., one of $Ucap X$, $Vcap X$ is empty. We conclude that $X$ is connected.
Assume $X$ is connected. Let $fcolon Xto Bbb R$ be continuous, $x_1,x_2in X$, $yin Bbb R$ with $f(x_1)<y<f(x_2)$. Then the sets $f^{-1}((-infty,y))$ and $f^{-1}((y,infty))$ are disjoint and, by continuity of $f$, open subsets of $X$, They are also non.empty (contain $x_1$ and $x_2$, respectively). By connectedness of $X$ they thus cannot cover $X$. Hence there must exist $xiin X$ such that $f(xi)notin(-infty,y)cup(y,infty)$, i.e., $f(xi)=y$.
Quite obviously, discrete subsets of $Bbb R$ (with at least two points) are not connected. Of course you can also see immediately that for any such discrete (or "gappy" in any other way) subset of $Bbb R$, the continuous map $xmapsto x$ breaks the Intermediate value prperty.
answered Jan 20 at 15:30
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
$begingroup$
Why can we assume that there exist two open sets $U, V$ such that $Xcap U$ and $Xcap V$ are disjoint? thanks in advance.
$endgroup$
– StammeringMathematician
Jan 20 at 16:02
add a comment |
$begingroup$
Let's say that a subset $Bbb X$ of $Bbb R$ has the IVT property if for every continuous map $fcolon Xto Bbb R$ and every $x_1,x_2in X$, $yin Bbb R$ with $f(x_1)<y<f(x_2)$, there exists $xiin X$ with $f(xi)=y$.
Then $X$ has the IVT property iff $X$ is connected.
Assume $X$ has the IVT property. Let $U,Vsubseteq Bbb R$ be open sets such that $Xsubseteq Ucup V$ and the sets $Xcap U$ and $Xcap V$ are disjoint. Define $fcolon XtoBbb R$ as $f(x)=begin{cases}1&xin U\0&text{otherwise}end{cases}$. Then $f$ is continuous. If there were $x_1,x_2in X$ with $f(x_1)=0$ and $f(x_2)=1$, then the IVT property would tell us that $f(xi)=frac12$ for some $xiin X$. As this is absurd, such $x_1,x_2$ cannot exist, i.e., one of $Ucap X$, $Vcap X$ is empty. We conclude that $X$ is connected.
Assume $X$ is connected. Let $fcolon Xto Bbb R$ be continuous, $x_1,x_2in X$, $yin Bbb R$ with $f(x_1)<y<f(x_2)$. Then the sets $f^{-1}((-infty,y))$ and $f^{-1}((y,infty))$ are disjoint and, by continuity of $f$, open subsets of $X$, They are also non.empty (contain $x_1$ and $x_2$, respectively). By connectedness of $X$ they thus cannot cover $X$. Hence there must exist $xiin X$ such that $f(xi)notin(-infty,y)cup(y,infty)$, i.e., $f(xi)=y$.
Quite obviously, discrete subsets of $Bbb R$ (with at least two points) are not connected. Of course you can also see immediately that for any such discrete (or "gappy" in any other way) subset of $Bbb R$, the continuous map $xmapsto x$ breaks the Intermediate value prperty.
answered Jan 20 at 15:30
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
Let's say that a subset $Bbb X$ of $Bbb R$ has the IVT property if for every continuous map $fcolon Xto Bbb R$ and every $x_1,x_2in X$, $yin Bbb R$ with $f(x_1)<y<f(x_2)$, there exists $xiin X$ with $f(xi)=y$.
Then $X$ has the IVT property iff $X$ is connected.
Assume $X$ has the IVT property. Let $U,Vsubseteq Bbb R$ be open sets such that $Xsubseteq Ucup V$ and the sets $Xcap U$ and $Xcap V$ are disjoint. Define $fcolon XtoBbb R$ as $f(x)=begin{cases}1&xin U\0&text{otherwise}end{cases}$. Then $f$ is continuous. If there were $x_1,x_2in X$ with $f(x_1)=0$ and $f(x_2)=1$, then the IVT property would tell us that $f(xi)=frac12$ for some $xiin X$. As this is absurd, such $x_1,x_2$ cannot exist, i.e., one of $Ucap X$, $Vcap X$ is empty. We conclude that $X$ is connected.
Assume $X$ is connected. Let $fcolon Xto Bbb R$ be continuous, $x_1,x_2in X$, $yin Bbb R$ with $f(x_1)<y<f(x_2)$. Then the sets $f^{-1}((-infty,y))$ and $f^{-1}((y,infty))$ are disjoint and, by continuity of $f$, open subsets of $X$, They are also non.empty (contain $x_1$ and $x_2$, respectively). By connectedness of $X$ they thus cannot cover $X$. Hence there must exist $xiin X$ such that $f(xi)notin(-infty,y)cup(y,infty)$, i.e., $f(xi)=y$.
Quite obviously, discrete subsets of $Bbb R$ (with at least two points) are not connected. Of course you can also see immediately that for any such discrete (or "gappy" in any other way) subset of $Bbb R$, the continuous map $xmapsto x$ breaks the Intermediate value prperty.
answered Jan 20 at 15:30
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 20 at 15:30
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 20 at 15:30
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 20 at 15:30
Hagen von EitzenHagen von Eitzen
282k23272505
282k23272505
$begingroup$
Why can we assume that there exist two open sets $U, V$ such that $Xcap U$ and $Xcap V$ are disjoint? thanks in advance.
$endgroup$
– StammeringMathematician
Jan 20 at 16:02
add a comment |
$begingroup$
Why can we assume that there exist two open sets $U, V$ such that $Xcap U$ and $Xcap V$ are disjoint? thanks in advance.
$endgroup$
– StammeringMathematician
Jan 20 at 16:02
$begingroup$
Why can we assume that there exist two open sets $U, V$ such that $Xcap U$ and $Xcap V$ are disjoint? thanks in advance.
$endgroup$
– StammeringMathematician
Jan 20 at 16:02
$begingroup$
Why can we assume that there exist two open sets $U, V$ such that $Xcap U$ and $Xcap V$ are disjoint? thanks in advance.
$endgroup$
– StammeringMathematician
Jan 20 at 16:02
add a comment |
$begingroup$
No. It is not true that every function from $mathbb Z$ into $mathbb R$ satisfies the intermediate value property (note that every function from $mathbb Z$ into $mathbb R$ is continuous), in spite of the fact that $mathbb Z$ is discrete and that it satisfies the least upper bound property.
answered Jan 20 at 15:15
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
all subsets of Z don't satisfy l.u.b?
$endgroup$
– Dis-integrating
Jan 20 at 16:00
$begingroup$
Yes they do: a non-emtpy subset of $mathbb Z$ with an upper bound always has a least upper bound.
$endgroup$
– José Carlos Santos
Jan 20 at 16:04
$begingroup$
oh right yes, we need that it has an upper bound in the proof anyway
$endgroup$
– Dis-integrating
Jan 20 at 16:05
add a comment |
$begingroup$
No. It is not true that every function from $mathbb Z$ into $mathbb R$ satisfies the intermediate value property (note that every function from $mathbb Z$ into $mathbb R$ is continuous), in spite of the fact that $mathbb Z$ is discrete and that it satisfies the least upper bound property.
answered Jan 20 at 15:15
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
all subsets of Z don't satisfy l.u.b?
$endgroup$
– Dis-integrating
Jan 20 at 16:00
$begingroup$
Yes they do: a non-emtpy subset of $mathbb Z$ with an upper bound always has a least upper bound.
$endgroup$
– José Carlos Santos
Jan 20 at 16:04
$begingroup$
oh right yes, we need that it has an upper bound in the proof anyway
$endgroup$
– Dis-integrating
Jan 20 at 16:05
add a comment |
$begingroup$
No. It is not true that every function from $mathbb Z$ into $mathbb R$ satisfies the intermediate value property (note that every function from $mathbb Z$ into $mathbb R$ is continuous), in spite of the fact that $mathbb Z$ is discrete and that it satisfies the least upper bound property.
answered Jan 20 at 15:15
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
No. It is not true that every function from $mathbb Z$ into $mathbb R$ satisfies the intermediate value property (note that every function from $mathbb Z$ into $mathbb R$ is continuous), in spite of the fact that $mathbb Z$ is discrete and that it satisfies the least upper bound property.
answered Jan 20 at 15:15
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 20 at 15:15
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 20 at 15:15
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 20 at 15:15
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
$begingroup$
all subsets of Z don't satisfy l.u.b?
$endgroup$
– Dis-integrating
Jan 20 at 16:00
$begingroup$
Yes they do: a non-emtpy subset of $mathbb Z$ with an upper bound always has a least upper bound.
$endgroup$
– José Carlos Santos
Jan 20 at 16:04
$begingroup$
oh right yes, we need that it has an upper bound in the proof anyway
$endgroup$
– Dis-integrating
Jan 20 at 16:05
add a comment |
$begingroup$
all subsets of Z don't satisfy l.u.b?
$endgroup$
– Dis-integrating
Jan 20 at 16:00
$begingroup$
Yes they do: a non-emtpy subset of $mathbb Z$ with an upper bound always has a least upper bound.
$endgroup$
– José Carlos Santos
Jan 20 at 16:04
$begingroup$
oh right yes, we need that it has an upper bound in the proof anyway
$endgroup$
– Dis-integrating
Jan 20 at 16:05
$begingroup$
all subsets of Z don't satisfy l.u.b?
$endgroup$
– Dis-integrating
Jan 20 at 16:00
$begingroup$
all subsets of Z don't satisfy l.u.b?
$endgroup$
– Dis-integrating
Jan 20 at 16:00
$begingroup$
Yes they do: a non-emtpy subset of $mathbb Z$ with an upper bound always has a least upper bound.
$endgroup$
– José Carlos Santos
Jan 20 at 16:04
$begingroup$
Yes they do: a non-emtpy subset of $mathbb Z$ with an upper bound always has a least upper bound.
$endgroup$
– José Carlos Santos
Jan 20 at 16:04
$begingroup$
oh right yes, we need that it has an upper bound in the proof anyway
$endgroup$
– Dis-integrating
Jan 20 at 16:05
$begingroup$
oh right yes, we need that it has an upper bound in the proof anyway
$endgroup$
– Dis-integrating
Jan 20 at 16:05
add a comment |
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Because of trivial examples such as $Bbb Zto Bbb R$, $xmapsto x$
$endgroup$
– Hagen von Eitzen
Jan 20 at 15:14