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continuity of the inverse in $0$ using “the pre image of an open set is open”.

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2

1

$begingroup$

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

share|cite|improve this question

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

2

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked Mar 17 at 21:33

Marine GalantinMarine Galantin

940419

$endgroup$

share|cite|improve this question

asked Mar 17 at 21:33

Marine GalantinMarine Galantin

940419

share|cite|improve this question

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

2 Answers
2

active

oldest

votes

4

$begingroup$

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.

share|cite|improve this answer

edited Mar 17 at 21:55

answered Mar 17 at 21:39

KlausKlaus

2,782113

$endgroup$

add a comment | 

2

$begingroup$

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...

share|cite|improve this answer

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

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4

$begingroup$

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.

share|cite|improve this answer

edited Mar 17 at 21:55

answered Mar 17 at 21:39

KlausKlaus

2,782113

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

edited Mar 17 at 21:55

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.

share|cite|improve this answer

edited Mar 17 at 21:55

answered Mar 17 at 21:39

KlausKlaus

2,782113

$endgroup$

share|cite|improve this answer

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

$begingroup$

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...

share|cite|improve this answer

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

2

$begingroup$

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...

share|cite|improve this answer

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...

share|cite|improve this answer

answered Mar 17 at 21:43

TheSilverDoeTheSilverDoe

4,692215

$endgroup$

share|cite|improve this answer

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