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continuity of the inverse in $0$ using “the pre image of an open set is open”.
Inverse image of Borel setContinuity on open and closed intervalsContinuity on open intervalHelp using theorem: A function between metric spaces is continuous if and only if for all open sets in the codomain the pre-image of the set is openContinuous inverse function theorem and open intervals?Inverse Image of Function Implies ContinuityContinuity on closed vs. open intervalsProving a set is open using pre image and continuity of a functionIs this an open/ closed set? (Pre image, continuity question)Intuitive Explanation of Continuity, Inverse Image Definition
$begingroup$
I'm focusing on $frac 1 x^2$, but any function with a "jump" or going to infinity at one point works.
I was trying to convince myself about the definition of continuity using the topological definition.
It says that for any open set, the pre image has to be open. So i was thinking that looking at the graph of an uncontinuous function should show me a problem. Nevertheless, visualy, I do not see anything.
Here, with $frac 1 x^2 $, if I take an open set (for example an open interval) on the y-axis, I can always get an open set on the x-axis...
what am I missing ?
I have the same problem for cases with jumps. So I found out that in the case where one side of the jump is closed (so the limit on one side of the gap exists, meaning the function is continuous on the left for example), I get the contradiction I was expecting ! If I take an open set where the upper bound lies right beneath the second side of the function which is left continuous, I get a closed interval ! (contradiction).
But in the case (like on the graph beneath), where the middle point is in a totally different place and so the function is neither continuous on the left nor on the right, I, again, can't find a problem...
Please, what's going on ?
I tried to be as clear as I could. If, nevertheless, I haven't been able to make it consistent, please feel free to ask question in the comment section.
real-analysis continuity
asked Mar 17 at 21:33
Marine GalantinMarine Galantin
940419
$endgroup$
|
show 1 more comment
$begingroup$
I'm focusing on $frac 1 x^2$, but any function with a "jump" or going to infinity at one point works.
I was trying to convince myself about the definition of continuity using the topological definition.
It says that for any open set, the pre image has to be open. So i was thinking that looking at the graph of an uncontinuous function should show me a problem. Nevertheless, visualy, I do not see anything.
Here, with $frac 1 x^2 $, if I take an open set (for example an open interval) on the y-axis, I can always get an open set on the x-axis...
what am I missing ?
I have the same problem for cases with jumps. So I found out that in the case where one side of the jump is closed (so the limit on one side of the gap exists, meaning the function is continuous on the left for example), I get the contradiction I was expecting ! If I take an open set where the upper bound lies right beneath the second side of the function which is left continuous, I get a closed interval ! (contradiction).
But in the case (like on the graph beneath), where the middle point is in a totally different place and so the function is neither continuous on the left nor on the right, I, again, can't find a problem...
Please, what's going on ?
I tried to be as clear as I could. If, nevertheless, I haven't been able to make it consistent, please feel free to ask question in the comment section.
real-analysis continuity
asked Mar 17 at 21:33
Marine GalantinMarine Galantin
940419
$endgroup$
$begingroup$
The function $f(x) =1/x^2$ isn't continuous at $x=0$. Otherwise, it is indeed continuous everywhere in $mathbbRsetminus0$.
$endgroup$
– Subhasis Biswas
Mar 17 at 21:39
$begingroup$
but how do you show that it isn't continuous? if for example you set $frac 1 x^2$ to be equal to $0$ in $0$
$endgroup$
– Marine Galantin
Mar 17 at 21:40
$begingroup$
As $x to 0$, the function becomes unbounded. If you define $f(0)=0$, It still fails to be continuous.
$endgroup$
– Subhasis Biswas
Mar 17 at 21:42
$begingroup$
that's exactly my question, how do you show this using the open sets ?
$endgroup$
– Marine Galantin
Mar 17 at 21:43
1
$begingroup$
exactly @SubhasisBiswas but I think someone just give me a satisfying answer. Thank you !
$endgroup$
– Marine Galantin
Mar 17 at 21:50
|
show 1 more comment
$begingroup$
I'm focusing on $frac 1 x^2$, but any function with a "jump" or going to infinity at one point works.
I was trying to convince myself about the definition of continuity using the topological definition.
It says that for any open set, the pre image has to be open. So i was thinking that looking at the graph of an uncontinuous function should show me a problem. Nevertheless, visualy, I do not see anything.
Here, with $frac 1 x^2 $, if I take an open set (for example an open interval) on the y-axis, I can always get an open set on the x-axis...
what am I missing ?
I have the same problem for cases with jumps. So I found out that in the case where one side of the jump is closed (so the limit on one side of the gap exists, meaning the function is continuous on the left for example), I get the contradiction I was expecting ! If I take an open set where the upper bound lies right beneath the second side of the function which is left continuous, I get a closed interval ! (contradiction).
But in the case (like on the graph beneath), where the middle point is in a totally different place and so the function is neither continuous on the left nor on the right, I, again, can't find a problem...
Please, what's going on ?
I tried to be as clear as I could. If, nevertheless, I haven't been able to make it consistent, please feel free to ask question in the comment section.
real-analysis continuity
asked Mar 17 at 21:33
Marine GalantinMarine Galantin
940419
$endgroup$
I'm focusing on $frac 1 x^2$, but any function with a "jump" or going to infinity at one point works.
I was trying to convince myself about the definition of continuity using the topological definition.
It says that for any open set, the pre image has to be open. So i was thinking that looking at the graph of an uncontinuous function should show me a problem. Nevertheless, visualy, I do not see anything.
Here, with $frac 1 x^2 $, if I take an open set (for example an open interval) on the y-axis, I can always get an open set on the x-axis...
what am I missing ?
I have the same problem for cases with jumps. So I found out that in the case where one side of the jump is closed (so the limit on one side of the gap exists, meaning the function is continuous on the left for example), I get the contradiction I was expecting ! If I take an open set where the upper bound lies right beneath the second side of the function which is left continuous, I get a closed interval ! (contradiction).
But in the case (like on the graph beneath), where the middle point is in a totally different place and so the function is neither continuous on the left nor on the right, I, again, can't find a problem...
Please, what's going on ?
I tried to be as clear as I could. If, nevertheless, I haven't been able to make it consistent, please feel free to ask question in the comment section.
real-analysis continuity
real-analysis continuity
asked Mar 17 at 21:33
Marine GalantinMarine Galantin
940419
asked Mar 17 at 21:33
Marine GalantinMarine Galantin
940419
asked Mar 17 at 21:33
Marine GalantinMarine Galantin
940419
asked Mar 17 at 21:33
Marine GalantinMarine Galantin
940419
asked Mar 17 at 21:33
Marine GalantinMarine Galantin
940419
940419
$begingroup$
The function $f(x) =1/x^2$ isn't continuous at $x=0$. Otherwise, it is indeed continuous everywhere in $mathbbRsetminus0$.
$endgroup$
– Subhasis Biswas
Mar 17 at 21:39
$begingroup$
but how do you show that it isn't continuous? if for example you set $frac 1 x^2$ to be equal to $0$ in $0$
$endgroup$
– Marine Galantin
Mar 17 at 21:40
$begingroup$
As $x to 0$, the function becomes unbounded. If you define $f(0)=0$, It still fails to be continuous.
$endgroup$
– Subhasis Biswas
Mar 17 at 21:42
$begingroup$
that's exactly my question, how do you show this using the open sets ?
$endgroup$
– Marine Galantin
Mar 17 at 21:43
1
$begingroup$
exactly @SubhasisBiswas but I think someone just give me a satisfying answer. Thank you !
$endgroup$
– Marine Galantin
Mar 17 at 21:50
|
show 1 more comment
$begingroup$
The function $f(x) =1/x^2$ isn't continuous at $x=0$. Otherwise, it is indeed continuous everywhere in $mathbbRsetminus0$.
$endgroup$
– Subhasis Biswas
Mar 17 at 21:39
$begingroup$
but how do you show that it isn't continuous? if for example you set $frac 1 x^2$ to be equal to $0$ in $0$
$endgroup$
– Marine Galantin
Mar 17 at 21:40
$begingroup$
As $x to 0$, the function becomes unbounded. If you define $f(0)=0$, It still fails to be continuous.
$endgroup$
– Subhasis Biswas
Mar 17 at 21:42
$begingroup$
that's exactly my question, how do you show this using the open sets ?
$endgroup$
– Marine Galantin
Mar 17 at 21:43
1
$begingroup$
exactly @SubhasisBiswas but I think someone just give me a satisfying answer. Thank you !
$endgroup$
– Marine Galantin
Mar 17 at 21:50
$begingroup$
The function $f(x) =1/x^2$ isn't continuous at $x=0$. Otherwise, it is indeed continuous everywhere in $mathbbRsetminus0$.
$endgroup$
– Subhasis Biswas
Mar 17 at 21:39
$begingroup$
The function $f(x) =1/x^2$ isn't continuous at $x=0$. Otherwise, it is indeed continuous everywhere in $mathbbRsetminus0$.
$endgroup$
– Subhasis Biswas
Mar 17 at 21:39
$begingroup$
but how do you show that it isn't continuous? if for example you set $frac 1 x^2$ to be equal to $0$ in $0$
$endgroup$
– Marine Galantin
Mar 17 at 21:40
$begingroup$
but how do you show that it isn't continuous? if for example you set $frac 1 x^2$ to be equal to $0$ in $0$
$endgroup$
– Marine Galantin
Mar 17 at 21:40
$begingroup$
As $x to 0$, the function becomes unbounded. If you define $f(0)=0$, It still fails to be continuous.
$endgroup$
– Subhasis Biswas
Mar 17 at 21:42
$begingroup$
As $x to 0$, the function becomes unbounded. If you define $f(0)=0$, It still fails to be continuous.
$endgroup$
– Subhasis Biswas
Mar 17 at 21:42
$begingroup$
that's exactly my question, how do you show this using the open sets ?
$endgroup$
– Marine Galantin
Mar 17 at 21:43
$begingroup$
that's exactly my question, how do you show this using the open sets ?
$endgroup$
– Marine Galantin
Mar 17 at 21:43
1
1
$begingroup$
exactly @SubhasisBiswas but I think someone just give me a satisfying answer. Thank you !
$endgroup$
– Marine Galantin
Mar 17 at 21:50
$begingroup$
exactly @SubhasisBiswas but I think someone just give me a satisfying answer. Thank you !
$endgroup$
– Marine Galantin
Mar 17 at 21:50
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The function $f: x mapsto frac1x^2$ is actually continuous as $0$ is not in the domain. If you set $f(0) = 0$, you get for example
$$f^-1((-1,1)) = (-infty,-1) cup (1,infty) cup 0,$$
which is not open.
answered Mar 17 at 21:39
KlausKlaus
2,782113
$endgroup$
add a comment |
$begingroup$
Actually, the $1/x^2$ function is not defined for $x=0$.
The natural way of defining it is to define it on $mathbbR setminus lbrace 0 rbrace$, on which it is continuous...
answered Mar 17 at 21:43
TheSilverDoeTheSilverDoe
4,692215
$endgroup$
$begingroup$
thank you for your answer but I think you're not getting the point. I'm trying to understand why if you define the value of this function at 0 for instance, this is still not continuous... as in the second part of my question. What you're saying is true, and I know it hahaha I'm studying math since a few years now. But I'm trying to understand continuity in another way.
$endgroup$
– Marine Galantin
Mar 17 at 21:45
$begingroup$
Ok. If you define $f(0)=a in mathbbR$, consider the open set $]a-1, a+1[$ of $mathbbR$. Can you see that the preimage of this set is not open ?
$endgroup$
– TheSilverDoe
Mar 17 at 21:47
$begingroup$
Yeah, I just saw that... in the case where the function is bounded but not continuous : either you choose to put the "a" on one side of the function making the function left/right continuous and so you're getting a closed interval, or you decide to put it somewhere else and you're getting a set of only one point which is closed. Either cases, you can find a pre image which is not open. Thank you !
$endgroup$
– Marine Galantin
Mar 17 at 21:49
$begingroup$
You get it ;) and you're welcome !
$endgroup$
– TheSilverDoe
Mar 17 at 21:50
$begingroup$
@TheSilverDoe, the inverse image set becomes a singleton set $0$, which is closed. Right?
$endgroup$
– Subhasis Biswas
Mar 17 at 21:52
|
show 1 more comment
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The function $f: x mapsto frac1x^2$ is actually continuous as $0$ is not in the domain. If you set $f(0) = 0$, you get for example
$$f^-1((-1,1)) = (-infty,-1) cup (1,infty) cup 0,$$
which is not open.
answered Mar 17 at 21:39
KlausKlaus
2,782113
$endgroup$
add a comment |
$begingroup$
The function $f: x mapsto frac1x^2$ is actually continuous as $0$ is not in the domain. If you set $f(0) = 0$, you get for example
$$f^-1((-1,1)) = (-infty,-1) cup (1,infty) cup 0,$$
which is not open.
answered Mar 17 at 21:39
KlausKlaus
2,782113
$endgroup$
add a comment |
$begingroup$
The function $f: x mapsto frac1x^2$ is actually continuous as $0$ is not in the domain. If you set $f(0) = 0$, you get for example
$$f^-1((-1,1)) = (-infty,-1) cup (1,infty) cup 0,$$
which is not open.
answered Mar 17 at 21:39
KlausKlaus
2,782113
$endgroup$
The function $f: x mapsto frac1x^2$ is actually continuous as $0$ is not in the domain. If you set $f(0) = 0$, you get for example
$$f^-1((-1,1)) = (-infty,-1) cup (1,infty) cup 0,$$
which is not open.
answered Mar 17 at 21:39
KlausKlaus
2,782113
edited Mar 17 at 21:55
answered Mar 17 at 21:39
KlausKlaus
2,782113
answered Mar 17 at 21:39
KlausKlaus
2,782113
answered Mar 17 at 21:39
KlausKlaus
2,782113
2,782113
add a comment |
add a comment |
$begingroup$
Actually, the $1/x^2$ function is not defined for $x=0$.
The natural way of defining it is to define it on $mathbbR setminus lbrace 0 rbrace$, on which it is continuous...
answered Mar 17 at 21:43
TheSilverDoeTheSilverDoe
4,692215
$endgroup$
$begingroup$
thank you for your answer but I think you're not getting the point. I'm trying to understand why if you define the value of this function at 0 for instance, this is still not continuous... as in the second part of my question. What you're saying is true, and I know it hahaha I'm studying math since a few years now. But I'm trying to understand continuity in another way.
$endgroup$
– Marine Galantin
Mar 17 at 21:45
$begingroup$
Ok. If you define $f(0)=a in mathbbR$, consider the open set $]a-1, a+1[$ of $mathbbR$. Can you see that the preimage of this set is not open ?
$endgroup$
– TheSilverDoe
Mar 17 at 21:47
$begingroup$
Yeah, I just saw that... in the case where the function is bounded but not continuous : either you choose to put the "a" on one side of the function making the function left/right continuous and so you're getting a closed interval, or you decide to put it somewhere else and you're getting a set of only one point which is closed. Either cases, you can find a pre image which is not open. Thank you !
$endgroup$
– Marine Galantin
Mar 17 at 21:49
$begingroup$
You get it ;) and you're welcome !
$endgroup$
– TheSilverDoe
Mar 17 at 21:50
$begingroup$
@TheSilverDoe, the inverse image set becomes a singleton set $0$, which is closed. Right?
$endgroup$
– Subhasis Biswas
Mar 17 at 21:52
|
show 1 more comment
$begingroup$
Actually, the $1/x^2$ function is not defined for $x=0$.
The natural way of defining it is to define it on $mathbbR setminus lbrace 0 rbrace$, on which it is continuous...
answered Mar 17 at 21:43
TheSilverDoeTheSilverDoe
4,692215
$endgroup$
$begingroup$
thank you for your answer but I think you're not getting the point. I'm trying to understand why if you define the value of this function at 0 for instance, this is still not continuous... as in the second part of my question. What you're saying is true, and I know it hahaha I'm studying math since a few years now. But I'm trying to understand continuity in another way.
$endgroup$
– Marine Galantin
Mar 17 at 21:45
$begingroup$
Ok. If you define $f(0)=a in mathbbR$, consider the open set $]a-1, a+1[$ of $mathbbR$. Can you see that the preimage of this set is not open ?
$endgroup$
– TheSilverDoe
Mar 17 at 21:47
$begingroup$
Yeah, I just saw that... in the case where the function is bounded but not continuous : either you choose to put the "a" on one side of the function making the function left/right continuous and so you're getting a closed interval, or you decide to put it somewhere else and you're getting a set of only one point which is closed. Either cases, you can find a pre image which is not open. Thank you !
$endgroup$
– Marine Galantin
Mar 17 at 21:49
$begingroup$
You get it ;) and you're welcome !
$endgroup$
– TheSilverDoe
Mar 17 at 21:50
$begingroup$
@TheSilverDoe, the inverse image set becomes a singleton set $0$, which is closed. Right?
$endgroup$
– Subhasis Biswas
Mar 17 at 21:52
|
show 1 more comment
$begingroup$
Actually, the $1/x^2$ function is not defined for $x=0$.
The natural way of defining it is to define it on $mathbbR setminus lbrace 0 rbrace$, on which it is continuous...
answered Mar 17 at 21:43
TheSilverDoeTheSilverDoe
4,692215
$endgroup$
Actually, the $1/x^2$ function is not defined for $x=0$.
The natural way of defining it is to define it on $mathbbR setminus lbrace 0 rbrace$, on which it is continuous...
answered Mar 17 at 21:43
TheSilverDoeTheSilverDoe
4,692215
answered Mar 17 at 21:43
TheSilverDoeTheSilverDoe
4,692215
answered Mar 17 at 21:43
TheSilverDoeTheSilverDoe
4,692215
answered Mar 17 at 21:43
TheSilverDoeTheSilverDoe
4,692215
4,692215
$begingroup$
thank you for your answer but I think you're not getting the point. I'm trying to understand why if you define the value of this function at 0 for instance, this is still not continuous... as in the second part of my question. What you're saying is true, and I know it hahaha I'm studying math since a few years now. But I'm trying to understand continuity in another way.
$endgroup$
– Marine Galantin
Mar 17 at 21:45
$begingroup$
Ok. If you define $f(0)=a in mathbbR$, consider the open set $]a-1, a+1[$ of $mathbbR$. Can you see that the preimage of this set is not open ?
$endgroup$
– TheSilverDoe
Mar 17 at 21:47
$begingroup$
Yeah, I just saw that... in the case where the function is bounded but not continuous : either you choose to put the "a" on one side of the function making the function left/right continuous and so you're getting a closed interval, or you decide to put it somewhere else and you're getting a set of only one point which is closed. Either cases, you can find a pre image which is not open. Thank you !
$endgroup$
– Marine Galantin
Mar 17 at 21:49
$begingroup$
You get it ;) and you're welcome !
$endgroup$
– TheSilverDoe
Mar 17 at 21:50
$begingroup$
@TheSilverDoe, the inverse image set becomes a singleton set $0$, which is closed. Right?
$endgroup$
– Subhasis Biswas
Mar 17 at 21:52
|
show 1 more comment
$begingroup$
thank you for your answer but I think you're not getting the point. I'm trying to understand why if you define the value of this function at 0 for instance, this is still not continuous... as in the second part of my question. What you're saying is true, and I know it hahaha I'm studying math since a few years now. But I'm trying to understand continuity in another way.
$endgroup$
– Marine Galantin
Mar 17 at 21:45
$begingroup$
Ok. If you define $f(0)=a in mathbbR$, consider the open set $]a-1, a+1[$ of $mathbbR$. Can you see that the preimage of this set is not open ?
$endgroup$
– TheSilverDoe
Mar 17 at 21:47
$begingroup$
Yeah, I just saw that... in the case where the function is bounded but not continuous : either you choose to put the "a" on one side of the function making the function left/right continuous and so you're getting a closed interval, or you decide to put it somewhere else and you're getting a set of only one point which is closed. Either cases, you can find a pre image which is not open. Thank you !
$endgroup$
– Marine Galantin
Mar 17 at 21:49
$begingroup$
You get it ;) and you're welcome !
$endgroup$
– TheSilverDoe
Mar 17 at 21:50
$begingroup$
@TheSilverDoe, the inverse image set becomes a singleton set $0$, which is closed. Right?
$endgroup$
– Subhasis Biswas
Mar 17 at 21:52
$begingroup$
thank you for your answer but I think you're not getting the point. I'm trying to understand why if you define the value of this function at 0 for instance, this is still not continuous... as in the second part of my question. What you're saying is true, and I know it hahaha I'm studying math since a few years now. But I'm trying to understand continuity in another way.
$endgroup$
– Marine Galantin
Mar 17 at 21:45
$begingroup$
thank you for your answer but I think you're not getting the point. I'm trying to understand why if you define the value of this function at 0 for instance, this is still not continuous... as in the second part of my question. What you're saying is true, and I know it hahaha I'm studying math since a few years now. But I'm trying to understand continuity in another way.
$endgroup$
– Marine Galantin
Mar 17 at 21:45
$begingroup$
Ok. If you define $f(0)=a in mathbbR$, consider the open set $]a-1, a+1[$ of $mathbbR$. Can you see that the preimage of this set is not open ?
$endgroup$
– TheSilverDoe
Mar 17 at 21:47
$begingroup$
Ok. If you define $f(0)=a in mathbbR$, consider the open set $]a-1, a+1[$ of $mathbbR$. Can you see that the preimage of this set is not open ?
$endgroup$
– TheSilverDoe
Mar 17 at 21:47
$begingroup$
Yeah, I just saw that... in the case where the function is bounded but not continuous : either you choose to put the "a" on one side of the function making the function left/right continuous and so you're getting a closed interval, or you decide to put it somewhere else and you're getting a set of only one point which is closed. Either cases, you can find a pre image which is not open. Thank you !
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– Marine Galantin
Mar 17 at 21:49
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Yeah, I just saw that... in the case where the function is bounded but not continuous : either you choose to put the "a" on one side of the function making the function left/right continuous and so you're getting a closed interval, or you decide to put it somewhere else and you're getting a set of only one point which is closed. Either cases, you can find a pre image which is not open. Thank you !
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– Marine Galantin
Mar 17 at 21:49
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You get it ;) and you're welcome !
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– TheSilverDoe
Mar 17 at 21:50
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You get it ;) and you're welcome !
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– TheSilverDoe
Mar 17 at 21:50
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@TheSilverDoe, the inverse image set becomes a singleton set $0$, which is closed. Right?
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– Subhasis Biswas
Mar 17 at 21:52
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@TheSilverDoe, the inverse image set becomes a singleton set $0$, which is closed. Right?
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– Subhasis Biswas
Mar 17 at 21:52
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The function $f(x) =1/x^2$ isn't continuous at $x=0$. Otherwise, it is indeed continuous everywhere in $mathbbRsetminus0$.
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– Subhasis Biswas
Mar 17 at 21:39
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but how do you show that it isn't continuous? if for example you set $frac 1 x^2$ to be equal to $0$ in $0$
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– Marine Galantin
Mar 17 at 21:40
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As $x to 0$, the function becomes unbounded. If you define $f(0)=0$, It still fails to be continuous.
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– Subhasis Biswas
Mar 17 at 21:42
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that's exactly my question, how do you show this using the open sets ?
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– Marine Galantin
Mar 17 at 21:43
1
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exactly @SubhasisBiswas but I think someone just give me a satisfying answer. Thank you !
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– Marine Galantin
Mar 17 at 21:50