Is the set $frac x^2 /(1+x^2) x$ connected in $mathbb R$? The 2019 Stack Overflow Developer Survey Results Are Inproving a set is not path-connectedThe set $A=(x, y)in mathbbR^2:$ is connectedHow to distinguish a connected set or a disconnected set?Connected components of a given spaceIs the following set connected?Is the closure of an open connected set polygonally connected?Countable set (discrete) function continuous?The graph of $f$ is a connected subset of $mathbbR^2$.Is the set path-connected?Suppose that the sets $A_1,A_2 subset mathbbR^n $ are connected and that they are not disjoint. Prove that $A_1 cup A_2$ is connected.
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Is the set $frac x^2 /(1+x^2) x$ connected in $mathbb R$?
The 2019 Stack Overflow Developer Survey Results Are Inproving a set is not path-connectedThe set $A=$ is connectedHow to distinguish a connected set or a disconnected set?Connected components of a given spaceIs the following set connected?Is the closure of an open connected set polygonally connected?Countable set (discrete) function continuous?The graph of $f$ is a connected subset of $mathbbR^2$.Is the set path-connected?Suppose that the sets $A_1,A_2 subset mathbbR^n $ are connected and that they are not disjoint. Prove that $A_1 cup A_2$ is connected.
$begingroup$
The set lies in the interval $[0,1)$ with points in $mathbb Q$ so it becomes a subset of $mathbb Q$, the former being not connected and so the set also becomes disconnected?
analysis
edited Mar 24 at 5:45
Rócherz
3,0263823
asked Mar 24 at 5:40
Van ThanaVan Thana
1
$endgroup$
add a comment |
$begingroup$
The set lies in the interval $[0,1)$ with points in $mathbb Q$ so it becomes a subset of $mathbb Q$, the former being not connected and so the set also becomes disconnected?
analysis
edited Mar 24 at 5:45
Rócherz
3,0263823
asked Mar 24 at 5:40
Van ThanaVan Thana
1
$endgroup$
$begingroup$
Connectedness is for sets. You are asking if a function is connected.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:59
$begingroup$
Is $x$ a positive integer? Are you asking if the range of the function is connected?
$endgroup$
– Kavi Rama Murthy
Mar 24 at 6:01
$begingroup$
Very confusing. Edit.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 24 at 14:57
$begingroup$
$frac x^2 /(1+x^2) x$ is a formula, not a set. Before we can answer your question, you need to explain exactly what set you are talking about.
$endgroup$
– Paul Sinclair
Mar 24 at 15:49
add a comment |
$begingroup$
The set lies in the interval $[0,1)$ with points in $mathbb Q$ so it becomes a subset of $mathbb Q$, the former being not connected and so the set also becomes disconnected?
analysis
edited Mar 24 at 5:45
Rócherz
3,0263823
asked Mar 24 at 5:40
Van ThanaVan Thana
1
$endgroup$
The set lies in the interval $[0,1)$ with points in $mathbb Q$ so it becomes a subset of $mathbb Q$, the former being not connected and so the set also becomes disconnected?
analysis
analysis
edited Mar 24 at 5:45
Rócherz
3,0263823
asked Mar 24 at 5:40
Van ThanaVan Thana
1
edited Mar 24 at 5:45
Rócherz
3,0263823
asked Mar 24 at 5:40
Van ThanaVan Thana
1
edited Mar 24 at 5:45
Rócherz
3,0263823
edited Mar 24 at 5:45
Rócherz
3,0263823
edited Mar 24 at 5:45
Rócherz
3,0263823
3,0263823
asked Mar 24 at 5:40
Van ThanaVan Thana
1
asked Mar 24 at 5:40
Van ThanaVan Thana
1
asked Mar 24 at 5:40
Van ThanaVan Thana
1
1
$begingroup$
Connectedness is for sets. You are asking if a function is connected.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:59
$begingroup$
Is $x$ a positive integer? Are you asking if the range of the function is connected?
$endgroup$
– Kavi Rama Murthy
Mar 24 at 6:01
$begingroup$
Very confusing. Edit.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 24 at 14:57
$begingroup$
$frac x^2 /(1+x^2) x$ is a formula, not a set. Before we can answer your question, you need to explain exactly what set you are talking about.
$endgroup$
– Paul Sinclair
Mar 24 at 15:49
add a comment |
$begingroup$
Connectedness is for sets. You are asking if a function is connected.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:59
$begingroup$
Is $x$ a positive integer? Are you asking if the range of the function is connected?
$endgroup$
– Kavi Rama Murthy
Mar 24 at 6:01
$begingroup$
Very confusing. Edit.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 24 at 14:57
$begingroup$
$frac x^2 /(1+x^2) x$ is a formula, not a set. Before we can answer your question, you need to explain exactly what set you are talking about.
$endgroup$
– Paul Sinclair
Mar 24 at 15:49
$begingroup$
Connectedness is for sets. You are asking if a function is connected.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:59
$begingroup$
Connectedness is for sets. You are asking if a function is connected.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:59
$begingroup$
Is $x$ a positive integer? Are you asking if the range of the function is connected?
$endgroup$
– Kavi Rama Murthy
Mar 24 at 6:01
$begingroup$
Is $x$ a positive integer? Are you asking if the range of the function is connected?
$endgroup$
– Kavi Rama Murthy
Mar 24 at 6:01
$begingroup$
Very confusing. Edit.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 24 at 14:57
$begingroup$
Very confusing. Edit.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 24 at 14:57
$begingroup$
$frac x^2 /(1+x^2) x$ is a formula, not a set. Before we can answer your question, you need to explain exactly what set you are talking about.
$endgroup$
– Paul Sinclair
Mar 24 at 15:49
$begingroup$
$frac x^2 /(1+x^2) x$ is a formula, not a set. Before we can answer your question, you need to explain exactly what set you are talking about.
$endgroup$
– Paul Sinclair
Mar 24 at 15:49
add a comment |
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$begingroup$
Connectedness is for sets. You are asking if a function is connected.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:59
$begingroup$
Is $x$ a positive integer? Are you asking if the range of the function is connected?
$endgroup$
– Kavi Rama Murthy
Mar 24 at 6:01
$begingroup$
Very confusing. Edit.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 24 at 14:57
$begingroup$
$frac x^2 /(1+x^2) x$ is a formula, not a set. Before we can answer your question, you need to explain exactly what set you are talking about.
$endgroup$
– Paul Sinclair
Mar 24 at 15:49