Counterexamples about function discontinuity.
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Let $f:mathbb{R}^nrightarrow mathbb{R}^n$ be a function with a point $textbf{x}inmathbb{R}^n$ of discontinuity. Is it possible that the image $f(O_{x_i})$, the image of an open ball (containing $x$) under $f$ to be connected for all $O_{x_i}$?
general-topology continuity examples-counterexamples connectedness
edited Dec 7 '18 at 19:28
José Carlos Santos
157k22126227
asked Dec 7 '18 at 19:13
William SunWilliam Sun
471111
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}^nrightarrow mathbb{R}^n$ be a function with a point $textbf{x}inmathbb{R}^n$ of discontinuity. Is it possible that the image $f(O_{x_i})$, the image of an open ball (containing $x$) under $f$ to be connected for all $O_{x_i}$?
general-topology continuity examples-counterexamples connectedness
edited Dec 7 '18 at 19:28
José Carlos Santos
157k22126227
asked Dec 7 '18 at 19:13
William SunWilliam Sun
471111
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}^nrightarrow mathbb{R}^n$ be a function with a point $textbf{x}inmathbb{R}^n$ of discontinuity. Is it possible that the image $f(O_{x_i})$, the image of an open ball (containing $x$) under $f$ to be connected for all $O_{x_i}$?
general-topology continuity examples-counterexamples connectedness
edited Dec 7 '18 at 19:28
José Carlos Santos
157k22126227
asked Dec 7 '18 at 19:13
William SunWilliam Sun
471111
$endgroup$
Let $f:mathbb{R}^nrightarrow mathbb{R}^n$ be a function with a point $textbf{x}inmathbb{R}^n$ of discontinuity. Is it possible that the image $f(O_{x_i})$, the image of an open ball (containing $x$) under $f$ to be connected for all $O_{x_i}$?
general-topology continuity examples-counterexamples connectedness
general-topology continuity examples-counterexamples connectedness
edited Dec 7 '18 at 19:28
José Carlos Santos
157k22126227
asked Dec 7 '18 at 19:13
William SunWilliam Sun
471111
edited Dec 7 '18 at 19:28
José Carlos Santos
157k22126227
asked Dec 7 '18 at 19:13
William SunWilliam Sun
471111
edited Dec 7 '18 at 19:28
José Carlos Santos
157k22126227
edited Dec 7 '18 at 19:28
José Carlos Santos
157k22126227
edited Dec 7 '18 at 19:28
José Carlos Santos
157k22126227
157k22126227
asked Dec 7 '18 at 19:13
William SunWilliam Sun
471111
asked Dec 7 '18 at 19:13
William SunWilliam Sun
471111
asked Dec 7 '18 at 19:13
William SunWilliam Sun
471111
471111
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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Yes. Take$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}sinleft(frac1xright)&text{ if }xneq0\0&text{ otherwise.}end{cases}end{array}$$Then $f$ is discontinuous at $0$, but the image of every open interval containing $0$ is $[-1,1]$, which is connected.
answered Dec 7 '18 at 19:27
José Carlos SantosJosé Carlos Santos
157k22126227
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add a comment |
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It is even possible that $f:mathbb Rto mathbb R$ maps every interval of positive length onto $mathbb R.$
answered Dec 7 '18 at 19:54
zhw.zhw.
72.4k43175
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Yes. Take$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}sinleft(frac1xright)&text{ if }xneq0\0&text{ otherwise.}end{cases}end{array}$$Then $f$ is discontinuous at $0$, but the image of every open interval containing $0$ is $[-1,1]$, which is connected.
answered Dec 7 '18 at 19:27
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
add a comment |
$begingroup$
Yes. Take$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}sinleft(frac1xright)&text{ if }xneq0\0&text{ otherwise.}end{cases}end{array}$$Then $f$ is discontinuous at $0$, but the image of every open interval containing $0$ is $[-1,1]$, which is connected.
answered Dec 7 '18 at 19:27
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
add a comment |
$begingroup$
Yes. Take$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}sinleft(frac1xright)&text{ if }xneq0\0&text{ otherwise.}end{cases}end{array}$$Then $f$ is discontinuous at $0$, but the image of every open interval containing $0$ is $[-1,1]$, which is connected.
answered Dec 7 '18 at 19:27
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
Yes. Take$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}sinleft(frac1xright)&text{ if }xneq0\0&text{ otherwise.}end{cases}end{array}$$Then $f$ is discontinuous at $0$, but the image of every open interval containing $0$ is $[-1,1]$, which is connected.
answered Dec 7 '18 at 19:27
José Carlos SantosJosé Carlos Santos
157k22126227
answered Dec 7 '18 at 19:27
José Carlos SantosJosé Carlos Santos
157k22126227
answered Dec 7 '18 at 19:27
José Carlos SantosJosé Carlos Santos
157k22126227
answered Dec 7 '18 at 19:27
José Carlos SantosJosé Carlos Santos
157k22126227
157k22126227
add a comment |
add a comment |
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It is even possible that $f:mathbb Rto mathbb R$ maps every interval of positive length onto $mathbb R.$
answered Dec 7 '18 at 19:54
zhw.zhw.
72.4k43175
$endgroup$
add a comment |
$begingroup$
It is even possible that $f:mathbb Rto mathbb R$ maps every interval of positive length onto $mathbb R.$
answered Dec 7 '18 at 19:54
zhw.zhw.
72.4k43175
$endgroup$
add a comment |
$begingroup$
It is even possible that $f:mathbb Rto mathbb R$ maps every interval of positive length onto $mathbb R.$
answered Dec 7 '18 at 19:54
zhw.zhw.
72.4k43175
$endgroup$
It is even possible that $f:mathbb Rto mathbb R$ maps every interval of positive length onto $mathbb R.$
answered Dec 7 '18 at 19:54
zhw.zhw.
72.4k43175
answered Dec 7 '18 at 19:54
zhw.zhw.
72.4k43175
answered Dec 7 '18 at 19:54
zhw.zhw.
72.4k43175
answered Dec 7 '18 at 19:54
zhw.zhw.
72.4k43175
72.4k43175
add a comment |
add a comment |
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