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minor question about Rudin, exercise 4.2 [closed]

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$begingroup$

Well, I solved the exercise and I had thoughts about it during the solving process. When I checked some solutions of it, it turned out to be a bit confusing, when making an assumption like this:
$f$ is a continuous mapping of a metric space $X$ and $E$ is a subset of $X$, let $x$ be a point of a closure of $E$. How can one make any assumptions about $f(x)$? For example when $E=X$ and some limit point of $E$ does not belong to $E$, then $f(x)$ is not even defined. So saying anything about $f(x)$ straightforwardly is not correct in this case. Am I wrong?

real-analysis

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asked Mar 20 at 11:33

shota kobakhidzeshota kobakhidze

165

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closed as off-topic by Thomas Shelby, Javi, José Carlos Santos, dantopa, clathratus Mar 20 at 20:11

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." – Thomas Shelby, Javi, dantopa

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Are we all supposed to know what that exercise is?
  $endgroup$
  – José Carlos Santos
  Mar 20 at 11:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, the exercise is not important itself but the conditions I wrote down and assumptions about $f(x)$ is what I am interested in.
  $endgroup$
  – shota kobakhidze
  Mar 20 at 11:36
  

  
  
  
  

add a comment | 

0

$begingroup$

Well, I solved the exercise and I had thoughts about it during the solving process. When I checked some solutions of it, it turned out to be a bit confusing, when making an assumption like this:
$f$ is a continuous mapping of a metric space $X$ and $E$ is a subset of $X$, let $x$ be a point of a closure of $E$. How can one make any assumptions about $f(x)$? For example when $E=X$ and some limit point of $E$ does not belong to $E$, then $f(x)$ is not even defined. So saying anything about $f(x)$ straightforwardly is not correct in this case. Am I wrong?

real-analysis

share|cite|improve this question

asked Mar 20 at 11:33

shota kobakhidzeshota kobakhidze

165

$endgroup$

closed as off-topic by Thomas Shelby, Javi, José Carlos Santos, dantopa, clathratus Mar 20 at 20:11

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." – Thomas Shelby, Javi, dantopa

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Are we all supposed to know what that exercise is?
  $endgroup$
  – José Carlos Santos
  Mar 20 at 11:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, the exercise is not important itself but the conditions I wrote down and assumptions about $f(x)$ is what I am interested in.
  $endgroup$
  – shota kobakhidze
  Mar 20 at 11:36
  

  
  
  
  

add a comment | 

$begingroup$

Well, I solved the exercise and I had thoughts about it during the solving process. When I checked some solutions of it, it turned out to be a bit confusing, when making an assumption like this:
$f$ is a continuous mapping of a metric space $X$ and $E$ is a subset of $X$, let $x$ be a point of a closure of $E$. How can one make any assumptions about $f(x)$? For example when $E=X$ and some limit point of $E$ does not belong to $E$, then $f(x)$ is not even defined. So saying anything about $f(x)$ straightforwardly is not correct in this case. Am I wrong?

real-analysis

share|cite|improve this question

asked Mar 20 at 11:33

shota kobakhidzeshota kobakhidze

165

$endgroup$

share|cite|improve this question

asked Mar 20 at 11:33

shota kobakhidzeshota kobakhidze

165

share|cite|improve this question

asked Mar 20 at 11:33

shota kobakhidzeshota kobakhidze

165

asked Mar 20 at 11:33

shota kobakhidzeshota kobakhidze

165

asked Mar 20 at 11:33

shota kobakhidzeshota kobakhidze

165

1 Answer
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1

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Look at definition $2.26$ in Rudin's book. It states that

If $X$ is a metric space, if $E subset X $, and if $E'$ denotes the set of all limit points of $E$ in $X$, then the closure of $E$ is the set $bar E = E cup E' $.

Since both $E$ and $E'$ are subsets of $X$, so is their union. If $f$ is defined on $X$, it is certainly defined on $E'$.

share|cite|improve this answer

edited Mar 20 at 11:52

answered Mar 20 at 11:49

JuliusL33tJuliusL33t

1,3681917

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, yeah. So it was a matter of definition. Thanks a lot!
  $endgroup$
  – shota kobakhidze
  Mar 20 at 11:50
  

  
  
  
  

add a comment | 

1

$begingroup$

Look at definition $2.26$ in Rudin's book. It states that

If $X$ is a metric space, if $E subset X $, and if $E'$ denotes the set of all limit points of $E$ in $X$, then the closure of $E$ is the set $bar E = E cup E' $.

Since both $E$ and $E'$ are subsets of $X$, so is their union. If $f$ is defined on $X$, it is certainly defined on $E'$.

share|cite|improve this answer

edited Mar 20 at 11:52

answered Mar 20 at 11:49

JuliusL33tJuliusL33t

1,3681917

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, yeah. So it was a matter of definition. Thanks a lot!
  $endgroup$
  – shota kobakhidze
  Mar 20 at 11:50
  

  
  
  
  

add a comment | 

1

$begingroup$

Look at definition $2.26$ in Rudin's book. It states that

If $X$ is a metric space, if $E subset X $, and if $E'$ denotes the set of all limit points of $E$ in $X$, then the closure of $E$ is the set $bar E = E cup E' $.

Since both $E$ and $E'$ are subsets of $X$, so is their union. If $f$ is defined on $X$, it is certainly defined on $E'$.

share|cite|improve this answer

edited Mar 20 at 11:52

answered Mar 20 at 11:49

JuliusL33tJuliusL33t

1,3681917

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, yeah. So it was a matter of definition. Thanks a lot!
  $endgroup$
  – shota kobakhidze
  Mar 20 at 11:50
  

  
  
  
  

add a comment | 

$begingroup$

Look at definition $2.26$ in Rudin's book. It states that

If $X$ is a metric space, if $E subset X $, and if $E'$ denotes the set of all limit points of $E$ in $X$, then the closure of $E$ is the set $bar E = E cup E' $.

Since both $E$ and $E'$ are subsets of $X$, so is their union. If $f$ is defined on $X$, it is certainly defined on $E'$.

share|cite|improve this answer

edited Mar 20 at 11:52

answered Mar 20 at 11:49

JuliusL33tJuliusL33t

1,3681917

$endgroup$

share|cite|improve this answer

edited Mar 20 at 11:52

answered Mar 20 at 11:49

JuliusL33tJuliusL33t

1,3681917

answered Mar 20 at 11:49

JuliusL33tJuliusL33t

1,3681917

answered Mar 20 at 11:49

JuliusL33tJuliusL33t

1,3681917