If f is increasing and g is continuous and for every q∈Q we have f(q)=g(q) , then f(x)=g(x) for every x∈R...
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I "can see" this proposition,but with the assumption that both functions are continuous..Surely i miss something.Any help?
real-analysis
asked Jan 12 at 16:31
Odu StavrouOdu Stavrou
72
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closed as off-topic by amWhy, RRL, Did, José Carlos Santos, Holo Jan 13 at 13:33
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If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
I "can see" this proposition,but with the assumption that both functions are continuous..Surely i miss something.Any help?
real-analysis
asked Jan 12 at 16:31
Odu StavrouOdu Stavrou
72
$endgroup$
closed as off-topic by amWhy, RRL, Did, José Carlos Santos, Holo Jan 13 at 13:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, RRL, Did, José Carlos Santos, Holo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I "can see" this proposition,but with the assumption that both functions are continuous..Surely i miss something.Any help?
real-analysis
asked Jan 12 at 16:31
Odu StavrouOdu Stavrou
72
$endgroup$
I "can see" this proposition,but with the assumption that both functions are continuous..Surely i miss something.Any help?
real-analysis
real-analysis
asked Jan 12 at 16:31
Odu StavrouOdu Stavrou
72
asked Jan 12 at 16:31
Odu StavrouOdu Stavrou
72
asked Jan 12 at 16:31
Odu StavrouOdu Stavrou
72
asked Jan 12 at 16:31
Odu StavrouOdu Stavrou
72
asked Jan 12 at 16:31
Odu StavrouOdu Stavrou
72
72
closed as off-topic by amWhy, RRL, Did, José Carlos Santos, Holo Jan 13 at 13:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, RRL, Did, José Carlos Santos, Holo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, RRL, Did, José Carlos Santos, Holo Jan 13 at 13:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, RRL, Did, José Carlos Santos, Holo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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1 Answer
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$begingroup$
Since $f$ is increasing, we can use the property $$
lim_{yuparrow x,;yinmathbb{Q}}f(y) le f(x) le lim_{ydownarrow x,;yin mathbb{Q}}f(y)
$$ for all $xinmathbb{R}$. Since $f=g$ on $mathbb{Q}$ and $g$ is continuous, we have
$$
g(x)=lim_{yuparrow x,;yinmathbb{Q}}f(y)le f(x)lelim_{ydownarrow x,;yin mathbb{Q}}f(y)=g(x).
$$
answered Jan 12 at 16:35
SongSong
10.8k628
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Probably worth mentioning that your property holds precisely because $f$ is increasing.
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– user3482749
Jan 12 at 16:36
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $f$ is increasing, we can use the property $$
lim_{yuparrow x,;yinmathbb{Q}}f(y) le f(x) le lim_{ydownarrow x,;yin mathbb{Q}}f(y)
$$ for all $xinmathbb{R}$. Since $f=g$ on $mathbb{Q}$ and $g$ is continuous, we have
$$
g(x)=lim_{yuparrow x,;yinmathbb{Q}}f(y)le f(x)lelim_{ydownarrow x,;yin mathbb{Q}}f(y)=g(x).
$$
answered Jan 12 at 16:35
SongSong
10.8k628
$endgroup$
$begingroup$
Probably worth mentioning that your property holds precisely because $f$ is increasing.
$endgroup$
– user3482749
Jan 12 at 16:36
add a comment |
$begingroup$
Since $f$ is increasing, we can use the property $$
lim_{yuparrow x,;yinmathbb{Q}}f(y) le f(x) le lim_{ydownarrow x,;yin mathbb{Q}}f(y)
$$ for all $xinmathbb{R}$. Since $f=g$ on $mathbb{Q}$ and $g$ is continuous, we have
$$
g(x)=lim_{yuparrow x,;yinmathbb{Q}}f(y)le f(x)lelim_{ydownarrow x,;yin mathbb{Q}}f(y)=g(x).
$$
answered Jan 12 at 16:35
SongSong
10.8k628
$endgroup$
$begingroup$
Probably worth mentioning that your property holds precisely because $f$ is increasing.
$endgroup$
– user3482749
Jan 12 at 16:36
add a comment |
$begingroup$
Since $f$ is increasing, we can use the property $$
lim_{yuparrow x,;yinmathbb{Q}}f(y) le f(x) le lim_{ydownarrow x,;yin mathbb{Q}}f(y)
$$ for all $xinmathbb{R}$. Since $f=g$ on $mathbb{Q}$ and $g$ is continuous, we have
$$
g(x)=lim_{yuparrow x,;yinmathbb{Q}}f(y)le f(x)lelim_{ydownarrow x,;yin mathbb{Q}}f(y)=g(x).
$$
answered Jan 12 at 16:35
SongSong
10.8k628
$endgroup$
Since $f$ is increasing, we can use the property $$
lim_{yuparrow x,;yinmathbb{Q}}f(y) le f(x) le lim_{ydownarrow x,;yin mathbb{Q}}f(y)
$$ for all $xinmathbb{R}$. Since $f=g$ on $mathbb{Q}$ and $g$ is continuous, we have
$$
g(x)=lim_{yuparrow x,;yinmathbb{Q}}f(y)le f(x)lelim_{ydownarrow x,;yin mathbb{Q}}f(y)=g(x).
$$
answered Jan 12 at 16:35
SongSong
10.8k628
edited Jan 12 at 16:36
answered Jan 12 at 16:35
SongSong
10.8k628
answered Jan 12 at 16:35
SongSong
10.8k628
answered Jan 12 at 16:35
SongSong
10.8k628
10.8k628
$begingroup$
Probably worth mentioning that your property holds precisely because $f$ is increasing.
$endgroup$
– user3482749
Jan 12 at 16:36
add a comment |
$begingroup$
Probably worth mentioning that your property holds precisely because $f$ is increasing.
$endgroup$
– user3482749
Jan 12 at 16:36
$begingroup$
Probably worth mentioning that your property holds precisely because $f$ is increasing.
$endgroup$
– user3482749
Jan 12 at 16:36
$begingroup$
Probably worth mentioning that your property holds precisely because $f$ is increasing.
$endgroup$
– user3482749
Jan 12 at 16:36
add a comment |
