Find a subset such that there is a cts function whose image is {0,1}
$begingroup$
Want to find a subset $X$ of $mathbb{R}^3$ such that there exists a cts function $f$ on the set such that the image $f(X) = {0,1}$, that is takes both values 0 and 1.
This is part of a larger question (IVT in higher dimensions) but just wanted to help understand this portion. Should be the case that some subsets of $mathbb{R}^3$ have such a function while others don't.
For a subset that doesn't, I came up with the set that consists only of the zero vector, ${(0,0,0)}$.
I might be overthinking it but struggling to come up with one since the function that exists needs to be continuous (otherwise I'd just do a piece wise function).
Appreciate any tips
real-analysis functions continuity
edited Nov 24 '18 at 1:36
Andrés E. Caicedo
65k8158246
asked Nov 24 '18 at 1:35
TommyTommy
1138
$endgroup$
add a comment |
$begingroup$
Want to find a subset $X$ of $mathbb{R}^3$ such that there exists a cts function $f$ on the set such that the image $f(X) = {0,1}$, that is takes both values 0 and 1.
This is part of a larger question (IVT in higher dimensions) but just wanted to help understand this portion. Should be the case that some subsets of $mathbb{R}^3$ have such a function while others don't.
For a subset that doesn't, I came up with the set that consists only of the zero vector, ${(0,0,0)}$.
I might be overthinking it but struggling to come up with one since the function that exists needs to be continuous (otherwise I'd just do a piece wise function).
Appreciate any tips
real-analysis functions continuity
edited Nov 24 '18 at 1:36
Andrés E. Caicedo
65k8158246
asked Nov 24 '18 at 1:35
TommyTommy
1138
$endgroup$
$begingroup$
Hint: the continuous image of a connected set is connected. So $X$ must be ....
$endgroup$
– user25959
Nov 24 '18 at 1:38
$begingroup$
So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
$endgroup$
– Tommy
Nov 24 '18 at 1:52
$begingroup$
Exactly - and check that the thing you find (use the simplest possible function) is continuous
$endgroup$
– user25959
Nov 24 '18 at 1:53
add a comment |
$begingroup$
Want to find a subset $X$ of $mathbb{R}^3$ such that there exists a cts function $f$ on the set such that the image $f(X) = {0,1}$, that is takes both values 0 and 1.
This is part of a larger question (IVT in higher dimensions) but just wanted to help understand this portion. Should be the case that some subsets of $mathbb{R}^3$ have such a function while others don't.
For a subset that doesn't, I came up with the set that consists only of the zero vector, ${(0,0,0)}$.
I might be overthinking it but struggling to come up with one since the function that exists needs to be continuous (otherwise I'd just do a piece wise function).
Appreciate any tips
real-analysis functions continuity
edited Nov 24 '18 at 1:36
Andrés E. Caicedo
65k8158246
asked Nov 24 '18 at 1:35
TommyTommy
1138
$endgroup$
Want to find a subset $X$ of $mathbb{R}^3$ such that there exists a cts function $f$ on the set such that the image $f(X) = {0,1}$, that is takes both values 0 and 1.
This is part of a larger question (IVT in higher dimensions) but just wanted to help understand this portion. Should be the case that some subsets of $mathbb{R}^3$ have such a function while others don't.
For a subset that doesn't, I came up with the set that consists only of the zero vector, ${(0,0,0)}$.
I might be overthinking it but struggling to come up with one since the function that exists needs to be continuous (otherwise I'd just do a piece wise function).
Appreciate any tips
real-analysis functions continuity
real-analysis functions continuity
edited Nov 24 '18 at 1:36
Andrés E. Caicedo
65k8158246
asked Nov 24 '18 at 1:35
TommyTommy
1138
edited Nov 24 '18 at 1:36
Andrés E. Caicedo
65k8158246
asked Nov 24 '18 at 1:35
TommyTommy
1138
edited Nov 24 '18 at 1:36
Andrés E. Caicedo
65k8158246
edited Nov 24 '18 at 1:36
Andrés E. Caicedo
65k8158246
edited Nov 24 '18 at 1:36
Andrés E. Caicedo
65k8158246
65k8158246
asked Nov 24 '18 at 1:35
TommyTommy
1138
asked Nov 24 '18 at 1:35
TommyTommy
1138
asked Nov 24 '18 at 1:35
TommyTommy
1138
1138
$begingroup$
Hint: the continuous image of a connected set is connected. So $X$ must be ....
$endgroup$
– user25959
Nov 24 '18 at 1:38
$begingroup$
So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
$endgroup$
– Tommy
Nov 24 '18 at 1:52
$begingroup$
Exactly - and check that the thing you find (use the simplest possible function) is continuous
$endgroup$
– user25959
Nov 24 '18 at 1:53
add a comment |
$begingroup$
Hint: the continuous image of a connected set is connected. So $X$ must be ....
$endgroup$
– user25959
Nov 24 '18 at 1:38
$begingroup$
So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
$endgroup$
– Tommy
Nov 24 '18 at 1:52
$begingroup$
Exactly - and check that the thing you find (use the simplest possible function) is continuous
$endgroup$
– user25959
Nov 24 '18 at 1:53
$begingroup$
Hint: the continuous image of a connected set is connected. So $X$ must be ....
$endgroup$
– user25959
Nov 24 '18 at 1:38
$begingroup$
Hint: the continuous image of a connected set is connected. So $X$ must be ....
$endgroup$
– user25959
Nov 24 '18 at 1:38
$begingroup$
So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
$endgroup$
– Tommy
Nov 24 '18 at 1:52
$begingroup$
So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
$endgroup$
– Tommy
Nov 24 '18 at 1:52
$begingroup$
Exactly - and check that the thing you find (use the simplest possible function) is continuous
$endgroup$
– user25959
Nov 24 '18 at 1:53
$begingroup$
Exactly - and check that the thing you find (use the simplest possible function) is continuous
$endgroup$
– user25959
Nov 24 '18 at 1:53
add a comment |
1 Answer
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$begingroup$
Hint: A crucial hypothesis in IVT is connectedness. Can you think of such a function on a disconnected domain?
answered Nov 24 '18 at 1:40
Zachary SelkZachary Selk
594311
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Hint: A crucial hypothesis in IVT is connectedness. Can you think of such a function on a disconnected domain?
answered Nov 24 '18 at 1:40
Zachary SelkZachary Selk
594311
$endgroup$
add a comment |
$begingroup$
Hint: A crucial hypothesis in IVT is connectedness. Can you think of such a function on a disconnected domain?
answered Nov 24 '18 at 1:40
Zachary SelkZachary Selk
594311
$endgroup$
add a comment |
$begingroup$
Hint: A crucial hypothesis in IVT is connectedness. Can you think of such a function on a disconnected domain?
answered Nov 24 '18 at 1:40
Zachary SelkZachary Selk
594311
$endgroup$
Hint: A crucial hypothesis in IVT is connectedness. Can you think of such a function on a disconnected domain?
answered Nov 24 '18 at 1:40
Zachary SelkZachary Selk
594311
answered Nov 24 '18 at 1:40
Zachary SelkZachary Selk
594311
answered Nov 24 '18 at 1:40
Zachary SelkZachary Selk
594311
answered Nov 24 '18 at 1:40
Zachary SelkZachary Selk
594311
594311
add a comment |
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$begingroup$
Hint: the continuous image of a connected set is connected. So $X$ must be ....
$endgroup$
– user25959
Nov 24 '18 at 1:38
$begingroup$
So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
$endgroup$
– Tommy
Nov 24 '18 at 1:52
$begingroup$
Exactly - and check that the thing you find (use the simplest possible function) is continuous
$endgroup$
– user25959
Nov 24 '18 at 1:53