How can we see that the proof of this following real analysis problems is reasonable?












5












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















5












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














5












5








5


2



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















7












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















7












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














7












7








7





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 20:29

























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


















  • $begingroup$
    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


















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