Mixing
On connected image of a disconnected domain
$begingroup$
Q.5 Let݂ $S=[0,1)cup [2,3]$ and $f:Sto R$ be a strictly increasing function such that $f(S)$ is connected. Which of the following statements is TRUE?
(A) ݂$f$ has exactly one discontinuity.
(B)݂$f$ has exactly two discontinuities.
(C)݂$f$ has infinitely many discontinuities.
(D)݂$f$ is continuous.
Can someone give some hints how to tackle this problem.
real-analysis general-topology
edited Jan 29 at 17:19
José Carlos Santos
171k23132240
asked Jan 29 at 17:18
user118413user118413
17210
$endgroup$
add a comment |
$begingroup$
Q.5 Let݂ $S=[0,1)cup [2,3]$ and $f:Sto R$ be a strictly increasing function such that $f(S)$ is connected. Which of the following statements is TRUE?
(A) ݂$f$ has exactly one discontinuity.
(B)݂$f$ has exactly two discontinuities.
(C)݂$f$ has infinitely many discontinuities.
(D)݂$f$ is continuous.
Can someone give some hints how to tackle this problem.
real-analysis general-topology
edited Jan 29 at 17:19
José Carlos Santos
171k23132240
asked Jan 29 at 17:18
user118413user118413
17210
$endgroup$
add a comment |
$begingroup$
Q.5 Let݂ $S=[0,1)cup [2,3]$ and $f:Sto R$ be a strictly increasing function such that $f(S)$ is connected. Which of the following statements is TRUE?
(A) ݂$f$ has exactly one discontinuity.
(B)݂$f$ has exactly two discontinuities.
(C)݂$f$ has infinitely many discontinuities.
(D)݂$f$ is continuous.
Can someone give some hints how to tackle this problem.
real-analysis general-topology
edited Jan 29 at 17:19
José Carlos Santos
171k23132240
asked Jan 29 at 17:18
user118413user118413
17210
$endgroup$
Q.5 Let݂ $S=[0,1)cup [2,3]$ and $f:Sto R$ be a strictly increasing function such that $f(S)$ is connected. Which of the following statements is TRUE?
(A) ݂$f$ has exactly one discontinuity.
(B)݂$f$ has exactly two discontinuities.
(C)݂$f$ has infinitely many discontinuities.
(D)݂$f$ is continuous.
Can someone give some hints how to tackle this problem.
real-analysis general-topology
real-analysis general-topology
edited Jan 29 at 17:19
José Carlos Santos
171k23132240
asked Jan 29 at 17:18
user118413user118413
17210
edited Jan 29 at 17:19
José Carlos Santos
171k23132240
asked Jan 29 at 17:18
user118413user118413
17210
edited Jan 29 at 17:19
José Carlos Santos
171k23132240
edited Jan 29 at 17:19
José Carlos Santos
171k23132240
edited Jan 29 at 17:19
José Carlos Santos
171k23132240
171k23132240
asked Jan 29 at 17:18
user118413user118413
17210
asked Jan 29 at 17:18
user118413user118413
17210
asked Jan 29 at 17:18
user118413user118413
17210
17210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hints:
The connected subsets of $mathbb{R}$ are intervals. (might be open, closed, half open, bounded, unbounded, it doesn't matter. They are intervals). How can you describe the image of $f$ then?
Because the function is increasing it has one sided limits at every point.
Try to combine both hints and it will get you to the solution.
answered Jan 29 at 17:31
MarkMark
10.4k1622
$endgroup$
$begingroup$
Therefore it has to be connected I think. The first components will map to some interval. As limit from left exist so the image of the second component will start exactly where the first component ends. Is it ok ?
$endgroup$
– user118413
Jan 29 at 17:35
$begingroup$
Good start. So what can you conclude about the continuity of the function?
$endgroup$
– Mark
Jan 29 at 17:37
$begingroup$
The function should be continuous.
$endgroup$
– user118413
Jan 29 at 17:40
$begingroup$
Yes, it is correct. Because the image is the interval $[f(0),f(3)]$ and the function is increasing it is easy to conclude that at each point $x$ the one sided limits must be equal to each other and $f(x)$ is a number between the values of these two limits. Because the one side limits are equal we get that $f(x)$ equals to the limit at the point $x$.
$endgroup$
– Mark
Jan 29 at 17:44
$begingroup$
Thank you so much for your kind attention
$endgroup$
– user118413
Jan 29 at 17:49
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hints:
The connected subsets of $mathbb{R}$ are intervals. (might be open, closed, half open, bounded, unbounded, it doesn't matter. They are intervals). How can you describe the image of $f$ then?
Because the function is increasing it has one sided limits at every point.
Try to combine both hints and it will get you to the solution.
answered Jan 29 at 17:31
MarkMark
10.4k1622
$endgroup$
$begingroup$
Therefore it has to be connected I think. The first components will map to some interval. As limit from left exist so the image of the second component will start exactly where the first component ends. Is it ok ?
$endgroup$
– user118413
Jan 29 at 17:35
$begingroup$
Good start. So what can you conclude about the continuity of the function?
$endgroup$
– Mark
Jan 29 at 17:37
$begingroup$
The function should be continuous.
$endgroup$
– user118413
Jan 29 at 17:40
$begingroup$
Yes, it is correct. Because the image is the interval $[f(0),f(3)]$ and the function is increasing it is easy to conclude that at each point $x$ the one sided limits must be equal to each other and $f(x)$ is a number between the values of these two limits. Because the one side limits are equal we get that $f(x)$ equals to the limit at the point $x$.
$endgroup$
– Mark
Jan 29 at 17:44
$begingroup$
Thank you so much for your kind attention
$endgroup$
– user118413
Jan 29 at 17:49
add a comment |
$begingroup$
Hints:
The connected subsets of $mathbb{R}$ are intervals. (might be open, closed, half open, bounded, unbounded, it doesn't matter. They are intervals). How can you describe the image of $f$ then?
Because the function is increasing it has one sided limits at every point.
Try to combine both hints and it will get you to the solution.
answered Jan 29 at 17:31
MarkMark
10.4k1622
$endgroup$
$begingroup$
Therefore it has to be connected I think. The first components will map to some interval. As limit from left exist so the image of the second component will start exactly where the first component ends. Is it ok ?
$endgroup$
– user118413
Jan 29 at 17:35
$begingroup$
Good start. So what can you conclude about the continuity of the function?
$endgroup$
– Mark
Jan 29 at 17:37
$begingroup$
The function should be continuous.
$endgroup$
– user118413
Jan 29 at 17:40
$begingroup$
Yes, it is correct. Because the image is the interval $[f(0),f(3)]$ and the function is increasing it is easy to conclude that at each point $x$ the one sided limits must be equal to each other and $f(x)$ is a number between the values of these two limits. Because the one side limits are equal we get that $f(x)$ equals to the limit at the point $x$.
$endgroup$
– Mark
Jan 29 at 17:44
$begingroup$
Thank you so much for your kind attention
$endgroup$
– user118413
Jan 29 at 17:49
add a comment |
$begingroup$
Hints:
The connected subsets of $mathbb{R}$ are intervals. (might be open, closed, half open, bounded, unbounded, it doesn't matter. They are intervals). How can you describe the image of $f$ then?
Because the function is increasing it has one sided limits at every point.
Try to combine both hints and it will get you to the solution.
answered Jan 29 at 17:31
MarkMark
10.4k1622
$endgroup$
Hints:
The connected subsets of $mathbb{R}$ are intervals. (might be open, closed, half open, bounded, unbounded, it doesn't matter. They are intervals). How can you describe the image of $f$ then?
Because the function is increasing it has one sided limits at every point.
Try to combine both hints and it will get you to the solution.
answered Jan 29 at 17:31
MarkMark
10.4k1622
answered Jan 29 at 17:31
MarkMark
10.4k1622
answered Jan 29 at 17:31
MarkMark
10.4k1622
answered Jan 29 at 17:31
MarkMark
10.4k1622
10.4k1622
$begingroup$
Therefore it has to be connected I think. The first components will map to some interval. As limit from left exist so the image of the second component will start exactly where the first component ends. Is it ok ?
$endgroup$
– user118413
Jan 29 at 17:35
$begingroup$
Good start. So what can you conclude about the continuity of the function?
$endgroup$
– Mark
Jan 29 at 17:37
$begingroup$
The function should be continuous.
$endgroup$
– user118413
Jan 29 at 17:40
$begingroup$
Yes, it is correct. Because the image is the interval $[f(0),f(3)]$ and the function is increasing it is easy to conclude that at each point $x$ the one sided limits must be equal to each other and $f(x)$ is a number between the values of these two limits. Because the one side limits are equal we get that $f(x)$ equals to the limit at the point $x$.
$endgroup$
– Mark
Jan 29 at 17:44
$begingroup$
Thank you so much for your kind attention
$endgroup$
– user118413
Jan 29 at 17:49
add a comment |
$begingroup$
Therefore it has to be connected I think. The first components will map to some interval. As limit from left exist so the image of the second component will start exactly where the first component ends. Is it ok ?
$endgroup$
– user118413
Jan 29 at 17:35
$begingroup$
Good start. So what can you conclude about the continuity of the function?
$endgroup$
– Mark
Jan 29 at 17:37
$begingroup$
The function should be continuous.
$endgroup$
– user118413
Jan 29 at 17:40
$begingroup$
Yes, it is correct. Because the image is the interval $[f(0),f(3)]$ and the function is increasing it is easy to conclude that at each point $x$ the one sided limits must be equal to each other and $f(x)$ is a number between the values of these two limits. Because the one side limits are equal we get that $f(x)$ equals to the limit at the point $x$.
$endgroup$
– Mark
Jan 29 at 17:44
$begingroup$
Thank you so much for your kind attention
$endgroup$
– user118413
Jan 29 at 17:49
$begingroup$
Therefore it has to be connected I think. The first components will map to some interval. As limit from left exist so the image of the second component will start exactly where the first component ends. Is it ok ?
$endgroup$
– user118413
Jan 29 at 17:35
$begingroup$
Therefore it has to be connected I think. The first components will map to some interval. As limit from left exist so the image of the second component will start exactly where the first component ends. Is it ok ?
$endgroup$
– user118413
Jan 29 at 17:35
$begingroup$
Good start. So what can you conclude about the continuity of the function?
$endgroup$
– Mark
Jan 29 at 17:37
$begingroup$
Good start. So what can you conclude about the continuity of the function?
$endgroup$
– Mark
Jan 29 at 17:37
$begingroup$
The function should be continuous.
$endgroup$
– user118413
Jan 29 at 17:40
$begingroup$
The function should be continuous.
$endgroup$
– user118413
Jan 29 at 17:40
$begingroup$
Yes, it is correct. Because the image is the interval $[f(0),f(3)]$ and the function is increasing it is easy to conclude that at each point $x$ the one sided limits must be equal to each other and $f(x)$ is a number between the values of these two limits. Because the one side limits are equal we get that $f(x)$ equals to the limit at the point $x$.
$endgroup$
– Mark
Jan 29 at 17:44
$begingroup$
Yes, it is correct. Because the image is the interval $[f(0),f(3)]$ and the function is increasing it is easy to conclude that at each point $x$ the one sided limits must be equal to each other and $f(x)$ is a number between the values of these two limits. Because the one side limits are equal we get that $f(x)$ equals to the limit at the point $x$.
$endgroup$
– Mark
Jan 29 at 17:44
$begingroup$
Thank you so much for your kind attention
$endgroup$
– user118413
Jan 29 at 17:49
$begingroup$
Thank you so much for your kind attention
$endgroup$
– user118413
Jan 29 at 17:49
add a comment |
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