How can we see that the proof of this following real analysis problems is reasonable?
$begingroup$
Problem: Let $f(x)$ be a real valued functions defined on $mathbb{R}$, prove that the point set $$E = {xin mathbb{R}: lim_{yrightarrow x} f(y) = +infty}$$ is a finite or a countable set.
Proof: Let $g(x) = arctan f(x), x in mathbb{R}$. Then the point set $E$ can be written as $$E = {xin mathbb{R}: lim_{yrightarrow x} g(y) = frac{pi}{2}}$$ Therefore $E$ is a finite or a countable set.
The above proof is from a textbook of real analysis. How can we see that the set $E$ in this proof is finite or countable?
real-analysis analysis lebesgue-measure
asked Jan 3 at 12:47
ScienceAgeScienceAge
713
$endgroup$
add a comment |
$begingroup$
Problem: Let $f(x)$ be a real valued functions defined on $mathbb{R}$, prove that the point set $$E = {xin mathbb{R}: lim_{yrightarrow x} f(y) = +infty}$$ is a finite or a countable set.
Proof: Let $g(x) = arctan f(x), x in mathbb{R}$. Then the point set $E$ can be written as $$E = {xin mathbb{R}: lim_{yrightarrow x} g(y) = frac{pi}{2}}$$ Therefore $E$ is a finite or a countable set.
The above proof is from a textbook of real analysis. How can we see that the set $E$ in this proof is finite or countable?
real-analysis analysis lebesgue-measure
asked Jan 3 at 12:47
ScienceAgeScienceAge
713
$endgroup$
2
$begingroup$
That's no proof at all! Labelling the two expressions in the obvious way, showing that $E_2$ is countable is no easier than showing that $E_1$ is countable. (Unless they've already proved the corresponding result for $f:Bbb Rto(a,b)$...)
$endgroup$
– David C. Ullrich
Jan 3 at 14:14
add a comment |
$begingroup$
Problem: Let $f(x)$ be a real valued functions defined on $mathbb{R}$, prove that the point set $$E = {xin mathbb{R}: lim_{yrightarrow x} f(y) = +infty}$$ is a finite or a countable set.
Proof: Let $g(x) = arctan f(x), x in mathbb{R}$. Then the point set $E$ can be written as $$E = {xin mathbb{R}: lim_{yrightarrow x} g(y) = frac{pi}{2}}$$ Therefore $E$ is a finite or a countable set.
The above proof is from a textbook of real analysis. How can we see that the set $E$ in this proof is finite or countable?
real-analysis analysis lebesgue-measure
asked Jan 3 at 12:47
ScienceAgeScienceAge
713
$endgroup$
Problem: Let $f(x)$ be a real valued functions defined on $mathbb{R}$, prove that the point set $$E = {xin mathbb{R}: lim_{yrightarrow x} f(y) = +infty}$$ is a finite or a countable set.
Proof: Let $g(x) = arctan f(x), x in mathbb{R}$. Then the point set $E$ can be written as $$E = {xin mathbb{R}: lim_{yrightarrow x} g(y) = frac{pi}{2}}$$ Therefore $E$ is a finite or a countable set.
The above proof is from a textbook of real analysis. How can we see that the set $E$ in this proof is finite or countable?
real-analysis analysis lebesgue-measure
real-analysis analysis lebesgue-measure
asked Jan 3 at 12:47
ScienceAgeScienceAge
713
asked Jan 3 at 12:47
ScienceAgeScienceAge
713
asked Jan 3 at 12:47
ScienceAgeScienceAge
713
asked Jan 3 at 12:47
ScienceAgeScienceAge
713
asked Jan 3 at 12:47
ScienceAgeScienceAge
713
713
2
$begingroup$
That's no proof at all! Labelling the two expressions in the obvious way, showing that $E_2$ is countable is no easier than showing that $E_1$ is countable. (Unless they've already proved the corresponding result for $f:Bbb Rto(a,b)$...)
$endgroup$
– David C. Ullrich
Jan 3 at 14:14
add a comment |
2
$begingroup$
That's no proof at all! Labelling the two expressions in the obvious way, showing that $E_2$ is countable is no easier than showing that $E_1$ is countable. (Unless they've already proved the corresponding result for $f:Bbb Rto(a,b)$...)
$endgroup$
– David C. Ullrich
Jan 3 at 14:14
2
2
$begingroup$
That's no proof at all! Labelling the two expressions in the obvious way, showing that $E_2$ is countable is no easier than showing that $E_1$ is countable. (Unless they've already proved the corresponding result for $f:Bbb Rto(a,b)$...)
$endgroup$
– David C. Ullrich
Jan 3 at 14:14
$begingroup$
That's no proof at all! Labelling the two expressions in the obvious way, showing that $E_2$ is countable is no easier than showing that $E_1$ is countable. (Unless they've already proved the corresponding result for $f:Bbb Rto(a,b)$...)
$endgroup$
– David C. Ullrich
Jan 3 at 14:14
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Right now, I have no idea why we can see immediately that $
E={ xin mathbb{R} ;|;lim_{yto x} g(y) = frac{pi}{2}}
$ is at most countable. But I've found the following argument that shows the set $E$ is at most countable.
Proof: Assume $x in E$. If $f(x)< n$, then since $lim_{yto x}f(y) =infty$, we can find a $delta>0$ such that
$$
(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty)).
$$ Now consider the open set $$U_n=text{int}f^{-1}((n,infty)) = bigcup_{k=1}^infty (alpha_k, beta_k)$$ where ${(alpha_k, beta_k)}$ is a disjoint family of open intervals. The above argument shows that $x$ is one of the end points $alpha_k$ or $beta_k$ of $U_n$ for some $nge 1$. Since there are at most countably many such end points, it follows that $E$ is countable.
$blacksquare$
After watching the above argument, I felt it is not trivial and it made me think that it is probably author's confusion, or there may be some results that the author did not mention.
answered Jan 3 at 14:15
SongSong
16.8k21145
$endgroup$
$begingroup$
Thanks, but why can we write $U_n=text{int}f^{-1}((n,infty))$ as a disjoint union of open intervals? This is what I am not clear about.
$endgroup$
– ScienceAge
Jan 5 at 7:49
1
$begingroup$
Oh, I was not so clear about that. Actually, it is just writing $U_n$ as a disjoint union of its components, and most topology textbooks perhaps would contain this proposition. If you are interested in the proof of it, you can see e.g. this earlier post math.stackexchange.com/questions/318299/….
$endgroup$
– Song
Jan 5 at 7:57
$begingroup$
Thanks. I wonder whether you think that if $(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty))$, then $(x-delta,x)cup (x,x+delta) subset text{int} f^{-1}((n,infty))$?
$endgroup$
– ScienceAge
Jan 6 at 3:24
$begingroup$
Yes, since the left-hand side is open in $mathbb{R}$, its interior is itself. Now we can take $text{int}$ operator both sides to see that it is true.
$endgroup$
– Song
Jan 6 at 3:48
add a comment |
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1 Answer
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$begingroup$
Right now, I have no idea why we can see immediately that $
E={ xin mathbb{R} ;|;lim_{yto x} g(y) = frac{pi}{2}}
$ is at most countable. But I've found the following argument that shows the set $E$ is at most countable.
Proof: Assume $x in E$. If $f(x)< n$, then since $lim_{yto x}f(y) =infty$, we can find a $delta>0$ such that
$$
(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty)).
$$ Now consider the open set $$U_n=text{int}f^{-1}((n,infty)) = bigcup_{k=1}^infty (alpha_k, beta_k)$$ where ${(alpha_k, beta_k)}$ is a disjoint family of open intervals. The above argument shows that $x$ is one of the end points $alpha_k$ or $beta_k$ of $U_n$ for some $nge 1$. Since there are at most countably many such end points, it follows that $E$ is countable.
$blacksquare$
After watching the above argument, I felt it is not trivial and it made me think that it is probably author's confusion, or there may be some results that the author did not mention.
answered Jan 3 at 14:15
SongSong
16.8k21145
$endgroup$
$begingroup$
Thanks, but why can we write $U_n=text{int}f^{-1}((n,infty))$ as a disjoint union of open intervals? This is what I am not clear about.
$endgroup$
– ScienceAge
Jan 5 at 7:49
1
$begingroup$
Oh, I was not so clear about that. Actually, it is just writing $U_n$ as a disjoint union of its components, and most topology textbooks perhaps would contain this proposition. If you are interested in the proof of it, you can see e.g. this earlier post math.stackexchange.com/questions/318299/….
$endgroup$
– Song
Jan 5 at 7:57
$begingroup$
Thanks. I wonder whether you think that if $(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty))$, then $(x-delta,x)cup (x,x+delta) subset text{int} f^{-1}((n,infty))$?
$endgroup$
– ScienceAge
Jan 6 at 3:24
$begingroup$
Yes, since the left-hand side is open in $mathbb{R}$, its interior is itself. Now we can take $text{int}$ operator both sides to see that it is true.
$endgroup$
– Song
Jan 6 at 3:48
add a comment |
$begingroup$
Right now, I have no idea why we can see immediately that $
E={ xin mathbb{R} ;|;lim_{yto x} g(y) = frac{pi}{2}}
$ is at most countable. But I've found the following argument that shows the set $E$ is at most countable.
Proof: Assume $x in E$. If $f(x)< n$, then since $lim_{yto x}f(y) =infty$, we can find a $delta>0$ such that
$$
(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty)).
$$ Now consider the open set $$U_n=text{int}f^{-1}((n,infty)) = bigcup_{k=1}^infty (alpha_k, beta_k)$$ where ${(alpha_k, beta_k)}$ is a disjoint family of open intervals. The above argument shows that $x$ is one of the end points $alpha_k$ or $beta_k$ of $U_n$ for some $nge 1$. Since there are at most countably many such end points, it follows that $E$ is countable.
$blacksquare$
After watching the above argument, I felt it is not trivial and it made me think that it is probably author's confusion, or there may be some results that the author did not mention.
answered Jan 3 at 14:15
SongSong
16.8k21145
$endgroup$
$begingroup$
Thanks, but why can we write $U_n=text{int}f^{-1}((n,infty))$ as a disjoint union of open intervals? This is what I am not clear about.
$endgroup$
– ScienceAge
Jan 5 at 7:49
1
$begingroup$
Oh, I was not so clear about that. Actually, it is just writing $U_n$ as a disjoint union of its components, and most topology textbooks perhaps would contain this proposition. If you are interested in the proof of it, you can see e.g. this earlier post math.stackexchange.com/questions/318299/….
$endgroup$
– Song
Jan 5 at 7:57
$begingroup$
Thanks. I wonder whether you think that if $(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty))$, then $(x-delta,x)cup (x,x+delta) subset text{int} f^{-1}((n,infty))$?
$endgroup$
– ScienceAge
Jan 6 at 3:24
$begingroup$
Yes, since the left-hand side is open in $mathbb{R}$, its interior is itself. Now we can take $text{int}$ operator both sides to see that it is true.
$endgroup$
– Song
Jan 6 at 3:48
add a comment |
$begingroup$
Right now, I have no idea why we can see immediately that $
E={ xin mathbb{R} ;|;lim_{yto x} g(y) = frac{pi}{2}}
$ is at most countable. But I've found the following argument that shows the set $E$ is at most countable.
Proof: Assume $x in E$. If $f(x)< n$, then since $lim_{yto x}f(y) =infty$, we can find a $delta>0$ such that
$$
(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty)).
$$ Now consider the open set $$U_n=text{int}f^{-1}((n,infty)) = bigcup_{k=1}^infty (alpha_k, beta_k)$$ where ${(alpha_k, beta_k)}$ is a disjoint family of open intervals. The above argument shows that $x$ is one of the end points $alpha_k$ or $beta_k$ of $U_n$ for some $nge 1$. Since there are at most countably many such end points, it follows that $E$ is countable.
$blacksquare$
After watching the above argument, I felt it is not trivial and it made me think that it is probably author's confusion, or there may be some results that the author did not mention.
answered Jan 3 at 14:15
SongSong
16.8k21145
$endgroup$
Right now, I have no idea why we can see immediately that $
E={ xin mathbb{R} ;|;lim_{yto x} g(y) = frac{pi}{2}}
$ is at most countable. But I've found the following argument that shows the set $E$ is at most countable.
Proof: Assume $x in E$. If $f(x)< n$, then since $lim_{yto x}f(y) =infty$, we can find a $delta>0$ such that
$$
(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty)).
$$ Now consider the open set $$U_n=text{int}f^{-1}((n,infty)) = bigcup_{k=1}^infty (alpha_k, beta_k)$$ where ${(alpha_k, beta_k)}$ is a disjoint family of open intervals. The above argument shows that $x$ is one of the end points $alpha_k$ or $beta_k$ of $U_n$ for some $nge 1$. Since there are at most countably many such end points, it follows that $E$ is countable.
$blacksquare$
After watching the above argument, I felt it is not trivial and it made me think that it is probably author's confusion, or there may be some results that the author did not mention.
answered Jan 3 at 14:15
SongSong
16.8k21145
edited Jan 3 at 20:29
answered Jan 3 at 14:15
SongSong
16.8k21145
answered Jan 3 at 14:15
SongSong
16.8k21145
answered Jan 3 at 14:15
SongSong
16.8k21145
16.8k21145
$begingroup$
Thanks, but why can we write $U_n=text{int}f^{-1}((n,infty))$ as a disjoint union of open intervals? This is what I am not clear about.
$endgroup$
– ScienceAge
Jan 5 at 7:49
1
$begingroup$
Oh, I was not so clear about that. Actually, it is just writing $U_n$ as a disjoint union of its components, and most topology textbooks perhaps would contain this proposition. If you are interested in the proof of it, you can see e.g. this earlier post math.stackexchange.com/questions/318299/….
$endgroup$
– Song
Jan 5 at 7:57
$begingroup$
Thanks. I wonder whether you think that if $(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty))$, then $(x-delta,x)cup (x,x+delta) subset text{int} f^{-1}((n,infty))$?
$endgroup$
– ScienceAge
Jan 6 at 3:24
$begingroup$
Yes, since the left-hand side is open in $mathbb{R}$, its interior is itself. Now we can take $text{int}$ operator both sides to see that it is true.
$endgroup$
– Song
Jan 6 at 3:48
add a comment |
$begingroup$
Thanks, but why can we write $U_n=text{int}f^{-1}((n,infty))$ as a disjoint union of open intervals? This is what I am not clear about.
$endgroup$
– ScienceAge
Jan 5 at 7:49
1
$begingroup$
Oh, I was not so clear about that. Actually, it is just writing $U_n$ as a disjoint union of its components, and most topology textbooks perhaps would contain this proposition. If you are interested in the proof of it, you can see e.g. this earlier post math.stackexchange.com/questions/318299/….
$endgroup$
– Song
Jan 5 at 7:57
$begingroup$
Thanks. I wonder whether you think that if $(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty))$, then $(x-delta,x)cup (x,x+delta) subset text{int} f^{-1}((n,infty))$?
$endgroup$
– ScienceAge
Jan 6 at 3:24
$begingroup$
Yes, since the left-hand side is open in $mathbb{R}$, its interior is itself. Now we can take $text{int}$ operator both sides to see that it is true.
$endgroup$
– Song
Jan 6 at 3:48
$begingroup$
Thanks, but why can we write $U_n=text{int}f^{-1}((n,infty))$ as a disjoint union of open intervals? This is what I am not clear about.
$endgroup$
– ScienceAge
Jan 5 at 7:49
$begingroup$
Thanks, but why can we write $U_n=text{int}f^{-1}((n,infty))$ as a disjoint union of open intervals? This is what I am not clear about.
$endgroup$
– ScienceAge
Jan 5 at 7:49
1
1
$begingroup$
Oh, I was not so clear about that. Actually, it is just writing $U_n$ as a disjoint union of its components, and most topology textbooks perhaps would contain this proposition. If you are interested in the proof of it, you can see e.g. this earlier post math.stackexchange.com/questions/318299/….
$endgroup$
– Song
Jan 5 at 7:57
$begingroup$
Oh, I was not so clear about that. Actually, it is just writing $U_n$ as a disjoint union of its components, and most topology textbooks perhaps would contain this proposition. If you are interested in the proof of it, you can see e.g. this earlier post math.stackexchange.com/questions/318299/….
$endgroup$
– Song
Jan 5 at 7:57
$begingroup$
Thanks. I wonder whether you think that if $(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty))$, then $(x-delta,x)cup (x,x+delta) subset text{int} f^{-1}((n,infty))$?
$endgroup$
– ScienceAge
Jan 6 at 3:24
$begingroup$
Thanks. I wonder whether you think that if $(x-delta,x)cup (x,x+delta) subset f^{-1}((n,infty))$, then $(x-delta,x)cup (x,x+delta) subset text{int} f^{-1}((n,infty))$?
$endgroup$
– ScienceAge
Jan 6 at 3:24
$begingroup$
Yes, since the left-hand side is open in $mathbb{R}$, its interior is itself. Now we can take $text{int}$ operator both sides to see that it is true.
$endgroup$
– Song
Jan 6 at 3:48
$begingroup$
Yes, since the left-hand side is open in $mathbb{R}$, its interior is itself. Now we can take $text{int}$ operator both sides to see that it is true.
$endgroup$
– Song
Jan 6 at 3:48
add a comment |
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$begingroup$
That's no proof at all! Labelling the two expressions in the obvious way, showing that $E_2$ is countable is no easier than showing that $E_1$ is countable. (Unless they've already proved the corresponding result for $f:Bbb Rto(a,b)$...)
$endgroup$
– David C. Ullrich
Jan 3 at 14:14