An Example of a Non-continuous Injection on a Real Interval, that is Not Strictly Monotonic.
up vote
1
down vote
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Theorem
Let $I$ be a real interval.
Let $f:I to mathbb R$ be an injective continuous real function.
Then $f$ is strictly monotone.
If the condition on continuity indeed is necessary (it most probably is but I would prefer to see it more clearly...), then we would be able to find such a function. Otherwise, I wouldn't be so convinced. I've seen the proof that uses the Intermediate Value Theorem and I do realize it requires continuity, obviously. But I don't know if we could do without it.
real-analysis continuity examples-counterexamples monotone-functions
asked Nov 17 at 12:46
FuzzyPixelz
337214
add a comment |
up vote
1
down vote
favorite
Theorem
Let $I$ be a real interval.
Let $f:I to mathbb R$ be an injective continuous real function.
Then $f$ is strictly monotone.
If the condition on continuity indeed is necessary (it most probably is but I would prefer to see it more clearly...), then we would be able to find such a function. Otherwise, I wouldn't be so convinced. I've seen the proof that uses the Intermediate Value Theorem and I do realize it requires continuity, obviously. But I don't know if we could do without it.
real-analysis continuity examples-counterexamples monotone-functions
asked Nov 17 at 12:46
FuzzyPixelz
337214
You could also use the mean value theorem to show the converse to that statement
– JB071098
Nov 17 at 12:51
1
But that isn't my question?
– FuzzyPixelz
Nov 17 at 12:53
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Theorem
Let $I$ be a real interval.
Let $f:I to mathbb R$ be an injective continuous real function.
Then $f$ is strictly monotone.
If the condition on continuity indeed is necessary (it most probably is but I would prefer to see it more clearly...), then we would be able to find such a function. Otherwise, I wouldn't be so convinced. I've seen the proof that uses the Intermediate Value Theorem and I do realize it requires continuity, obviously. But I don't know if we could do without it.
real-analysis continuity examples-counterexamples monotone-functions
asked Nov 17 at 12:46
FuzzyPixelz
337214
Theorem
Let $I$ be a real interval.
Let $f:I to mathbb R$ be an injective continuous real function.
Then $f$ is strictly monotone.
If the condition on continuity indeed is necessary (it most probably is but I would prefer to see it more clearly...), then we would be able to find such a function. Otherwise, I wouldn't be so convinced. I've seen the proof that uses the Intermediate Value Theorem and I do realize it requires continuity, obviously. But I don't know if we could do without it.
real-analysis continuity examples-counterexamples monotone-functions
real-analysis continuity examples-counterexamples monotone-functions
asked Nov 17 at 12:46
FuzzyPixelz
337214
asked Nov 17 at 12:46
FuzzyPixelz
337214
asked Nov 17 at 12:46
FuzzyPixelz
337214
asked Nov 17 at 12:46
FuzzyPixelz
337214
asked Nov 17 at 12:46
FuzzyPixelz
337214
337214
You could also use the mean value theorem to show the converse to that statement
– JB071098
Nov 17 at 12:51
1
But that isn't my question?
– FuzzyPixelz
Nov 17 at 12:53
add a comment |
You could also use the mean value theorem to show the converse to that statement
– JB071098
Nov 17 at 12:51
1
But that isn't my question?
– FuzzyPixelz
Nov 17 at 12:53
You could also use the mean value theorem to show the converse to that statement
– JB071098
Nov 17 at 12:51
You could also use the mean value theorem to show the converse to that statement
– JB071098
Nov 17 at 12:51
1
1
But that isn't my question?
– FuzzyPixelz
Nov 17 at 12:53
But that isn't my question?
– FuzzyPixelz
Nov 17 at 12:53
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Consider $I = [0,2]$ and $f$ defined on $I$ via $f(x) := x$ when $x leq 1$ and $f(x):= -x $ when $x > 1$.
answered Nov 17 at 13:11
S. van Nigtevecht
1,6751322
Well, I really feel silly right now... Thank you.
– FuzzyPixelz
Nov 17 at 13:23
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Consider $I = [0,2]$ and $f$ defined on $I$ via $f(x) := x$ when $x leq 1$ and $f(x):= -x $ when $x > 1$.
answered Nov 17 at 13:11
S. van Nigtevecht
1,6751322
Well, I really feel silly right now... Thank you.
– FuzzyPixelz
Nov 17 at 13:23
add a comment |
up vote
4
down vote
accepted
Consider $I = [0,2]$ and $f$ defined on $I$ via $f(x) := x$ when $x leq 1$ and $f(x):= -x $ when $x > 1$.
answered Nov 17 at 13:11
S. van Nigtevecht
1,6751322
Well, I really feel silly right now... Thank you.
– FuzzyPixelz
Nov 17 at 13:23
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Consider $I = [0,2]$ and $f$ defined on $I$ via $f(x) := x$ when $x leq 1$ and $f(x):= -x $ when $x > 1$.
answered Nov 17 at 13:11
S. van Nigtevecht
1,6751322
Consider $I = [0,2]$ and $f$ defined on $I$ via $f(x) := x$ when $x leq 1$ and $f(x):= -x $ when $x > 1$.
answered Nov 17 at 13:11
S. van Nigtevecht
1,6751322
answered Nov 17 at 13:11
S. van Nigtevecht
1,6751322
answered Nov 17 at 13:11
S. van Nigtevecht
1,6751322
answered Nov 17 at 13:11
S. van Nigtevecht
1,6751322
1,6751322
Well, I really feel silly right now... Thank you.
– FuzzyPixelz
Nov 17 at 13:23
add a comment |
Well, I really feel silly right now... Thank you.
– FuzzyPixelz
Nov 17 at 13:23
Well, I really feel silly right now... Thank you.
– FuzzyPixelz
Nov 17 at 13:23
Well, I really feel silly right now... Thank you.
– FuzzyPixelz
Nov 17 at 13:23
add a comment |
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You could also use the mean value theorem to show the converse to that statement
– JB071098
Nov 17 at 12:51
1
But that isn't my question?
– FuzzyPixelz
Nov 17 at 12:53