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How to approach finding the closure of a set

Calculate the closure and the interior of this setFormally prove the closure and interior of this setWhat is the closure of a circle under the order topology?Every open neighborhood of a point in the closure contains a point in the setIs a closed convex set $E$ in $mathbb{R}^n$ equal to the closure of its interior?The closure of $(0,1)$ in the lower-limit topology on $mathbb{R}$Let $A subset X$. Show that if $C$ is a closed set of $X$ and $C$ contains $A$, then $C$ contains the closure of $A$.Find the closure in $(mathbb{R}^2, tau)$ of the following sets.Prove the closure is closed and is contained in every closed setClosure of the set

2

1

$begingroup$

All I have found are various definitions of the clousure of a set. The only solving strategy I found was to intersect all closed subsets containing the subset whose closure I am looking for. However this does not help me out in my exercise in which I am asked to find the closure of $A$:

$$
A:=(Q cap (0,infty)) setminusleft{n^{-1} mid n in mathbb{N} right} text{ in } (0,infty)
$$

I have literally no idea how to approach this problem and did not find anything specific on the internet. Can someone please help me solving this?

real-analysis general-topology

share|cite|improve this question

edited 16 hours ago

postmortes

2,10031222

asked 17 hours ago

minitsminits

111

New contributor

minits is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
  $endgroup$
  – José Carlos Santos
  17 hours ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

All I have found are various definitions of the clousure of a set. The only solving strategy I found was to intersect all closed subsets containing the subset whose closure I am looking for. However this does not help me out in my exercise in which I am asked to find the closure of $A$:

$$
A:=(Q cap (0,infty)) setminusleft{n^{-1} mid n in mathbb{N} right} text{ in } (0,infty)
$$

I have literally no idea how to approach this problem and did not find anything specific on the internet. Can someone please help me solving this?

real-analysis general-topology

share|cite|improve this question

edited 16 hours ago

postmortes

2,10031222

asked 17 hours ago

minitsminits

111

New contributor

minits is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
  $endgroup$
  – José Carlos Santos
  17 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

All I have found are various definitions of the clousure of a set. The only solving strategy I found was to intersect all closed subsets containing the subset whose closure I am looking for. However this does not help me out in my exercise in which I am asked to find the closure of $A$:

$$
A:=(Q cap (0,infty)) setminusleft{n^{-1} mid n in mathbb{N} right} text{ in } (0,infty)
$$

I have literally no idea how to approach this problem and did not find anything specific on the internet. Can someone please help me solving this?

real-analysis general-topology

share|cite|improve this question

edited 16 hours ago

postmortes

2,10031222

asked 17 hours ago

minitsminits

111

New contributor

minits is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 16 hours ago

postmortes

2,10031222

asked 17 hours ago

minitsminits

111

New contributor

minits is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 16 hours ago

postmortes

2,10031222

asked 17 hours ago

minitsminits

111

New contributor

minits is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 16 hours ago

postmortes

2,10031222

edited 16 hours ago

postmortes

2,10031222

asked 17 hours ago

minitsminits

111

New contributor

minits is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 17 hours ago

minitsminits

111

1 Answer
1

active

oldest

votes

0

$begingroup$

Let $x in (0, +infty)$. Consider the sequence
$$u_n = frac{lfloor 10^nx rfloor}{10^n}$$

where $lfloor. rfloor$ denotes the floor function. You have
$$x- frac{1}{10^n} =frac{10^nx-1}{10^n} leq u_n leq frac{10^nx}{x} = x$$

so by comparison the sequence $(u_n)$ converges to $x$.

Now consider, for each $k$, consider the set $$A_k = leftlbrace n in mathbb{N}^* quad | quad frac{1}{n} in left(u_k, u_k + frac{1}{k} right) rightrbrace $$

This set is a subset of $mathbb{N}^*$, and therefore is either empty, either has a minimal element.

If $A_k$ is empty, let's define $v_k = u_k$.
If $A_k$ has a minimal element $n_0$, then define
$$v_k = max left( frac{u_k + frac{1}{n_0}}{2}, frac{frac{1}{n_0+1} + frac{1}{n_0}}{2} right)$$

I let you check that the sequence $(v_k)$ tends to $x$ and is composed of elements of $A$.

This would prove that the closure of $A$ is the entire $(0, +infty)$.

share|cite|improve this answer

answered 16 hours ago

TheSilverDoeTheSilverDoe

3,011112

$endgroup$

add a comment | 

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0

$begingroup$

Let $x in (0, +infty)$. Consider the sequence
$$u_n = frac{lfloor 10^nx rfloor}{10^n}$$

where $lfloor. rfloor$ denotes the floor function. You have
$$x- frac{1}{10^n} =frac{10^nx-1}{10^n} leq u_n leq frac{10^nx}{x} = x$$

so by comparison the sequence $(u_n)$ converges to $x$.

Now consider, for each $k$, consider the set $$A_k = leftlbrace n in mathbb{N}^* quad | quad frac{1}{n} in left(u_k, u_k + frac{1}{k} right) rightrbrace $$

This set is a subset of $mathbb{N}^*$, and therefore is either empty, either has a minimal element.

If $A_k$ is empty, let's define $v_k = u_k$.
If $A_k$ has a minimal element $n_0$, then define
$$v_k = max left( frac{u_k + frac{1}{n_0}}{2}, frac{frac{1}{n_0+1} + frac{1}{n_0}}{2} right)$$

I let you check that the sequence $(v_k)$ tends to $x$ and is composed of elements of $A$.

This would prove that the closure of $A$ is the entire $(0, +infty)$.

share|cite|improve this answer

answered 16 hours ago

TheSilverDoeTheSilverDoe

3,011112

$endgroup$

add a comment | 

0

$begingroup$

Let $x in (0, +infty)$. Consider the sequence
$$u_n = frac{lfloor 10^nx rfloor}{10^n}$$

where $lfloor. rfloor$ denotes the floor function. You have
$$x- frac{1}{10^n} =frac{10^nx-1}{10^n} leq u_n leq frac{10^nx}{x} = x$$

so by comparison the sequence $(u_n)$ converges to $x$.

Now consider, for each $k$, consider the set $$A_k = leftlbrace n in mathbb{N}^* quad | quad frac{1}{n} in left(u_k, u_k + frac{1}{k} right) rightrbrace $$

This set is a subset of $mathbb{N}^*$, and therefore is either empty, either has a minimal element.

If $A_k$ is empty, let's define $v_k = u_k$.
If $A_k$ has a minimal element $n_0$, then define
$$v_k = max left( frac{u_k + frac{1}{n_0}}{2}, frac{frac{1}{n_0+1} + frac{1}{n_0}}{2} right)$$

I let you check that the sequence $(v_k)$ tends to $x$ and is composed of elements of $A$.

This would prove that the closure of $A$ is the entire $(0, +infty)$.

share|cite|improve this answer

answered 16 hours ago

TheSilverDoeTheSilverDoe

3,011112

$endgroup$

add a comment | 

$begingroup$

Let $x in (0, +infty)$. Consider the sequence
$$u_n = frac{lfloor 10^nx rfloor}{10^n}$$

where $lfloor. rfloor$ denotes the floor function. You have
$$x- frac{1}{10^n} =frac{10^nx-1}{10^n} leq u_n leq frac{10^nx}{x} = x$$

so by comparison the sequence $(u_n)$ converges to $x$.

Now consider, for each $k$, consider the set $$A_k = leftlbrace n in mathbb{N}^* quad | quad frac{1}{n} in left(u_k, u_k + frac{1}{k} right) rightrbrace $$

This set is a subset of $mathbb{N}^*$, and therefore is either empty, either has a minimal element.

If $A_k$ is empty, let's define $v_k = u_k$.
If $A_k$ has a minimal element $n_0$, then define
$$v_k = max left( frac{u_k + frac{1}{n_0}}{2}, frac{frac{1}{n_0+1} + frac{1}{n_0}}{2} right)$$

I let you check that the sequence $(v_k)$ tends to $x$ and is composed of elements of $A$.

This would prove that the closure of $A$ is the entire $(0, +infty)$.

share|cite|improve this answer

answered 16 hours ago

TheSilverDoeTheSilverDoe

3,011112

$endgroup$

share|cite|improve this answer

answered 16 hours ago

TheSilverDoeTheSilverDoe

3,011112

answered 16 hours ago

TheSilverDoeTheSilverDoe

3,011112

answered 16 hours ago

TheSilverDoeTheSilverDoe

3,011112