Counter example of continous function such that there is Set S with $f(S)^circ subset (f(S^circ))$
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$S^circ$ denotes the interior of a set $S$. Is there an example of a continuous function $f$ and a set S with $(f(X))^circ notsubset f(S^circ)$ ?
I know that$f(S^circ)subset (f(S))^circ$ is not always true; for example
$$f(x)=x, ....[0,1]$$
$$f(x)=x-1 ....[2,3]$$
I tried hard but I could not find counterexample for $(f(S))^circsubset f(S^circ)$
Any help will be appreciated
real-analysis examples-counterexamples
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add a comment |
$begingroup$
$S^circ$ denotes the interior of a set $S$. Is there an example of a continuous function $f$ and a set S with $(f(X))^circ notsubset f(S^circ)$ ?
I know that$f(S^circ)subset (f(S))^circ$ is not always true; for example
$$f(x)=x, ....[0,1]$$
$$f(x)=x-1 ....[2,3]$$
I tried hard but I could not find counterexample for $(f(S))^circsubset f(S^circ)$
Any help will be appreciated
real-analysis examples-counterexamples
$endgroup$
add a comment |
$begingroup$
$S^circ$ denotes the interior of a set $S$. Is there an example of a continuous function $f$ and a set S with $(f(X))^circ notsubset f(S^circ)$ ?
I know that$f(S^circ)subset (f(S))^circ$ is not always true; for example
$$f(x)=x, ....[0,1]$$
$$f(x)=x-1 ....[2,3]$$
I tried hard but I could not find counterexample for $(f(S))^circsubset f(S^circ)$
Any help will be appreciated
real-analysis examples-counterexamples
$endgroup$
$S^circ$ denotes the interior of a set $S$. Is there an example of a continuous function $f$ and a set S with $(f(X))^circ notsubset f(S^circ)$ ?
I know that$f(S^circ)subset (f(S))^circ$ is not always true; for example
$$f(x)=x, ....[0,1]$$
$$f(x)=x-1 ....[2,3]$$
I tried hard but I could not find counterexample for $(f(S))^circsubset f(S^circ)$
Any help will be appreciated
real-analysis examples-counterexamples
real-analysis examples-counterexamples
edited Dec 6 '18 at 18:25
user25959
1,573816
1,573816
asked Dec 6 '18 at 16:30
MathLoverMathLover
49310
49310
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$begingroup$
This may be overkill, but the cantor function with $S=$ the middle-thirds cantor set will work:
2 properties of this function are: $S^circ = emptyset$ and $f(S)=[0,1]$
$endgroup$
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$begingroup$
This may be overkill, but the cantor function with $S=$ the middle-thirds cantor set will work:
2 properties of this function are: $S^circ = emptyset$ and $f(S)=[0,1]$
$endgroup$
add a comment |
$begingroup$
This may be overkill, but the cantor function with $S=$ the middle-thirds cantor set will work:
2 properties of this function are: $S^circ = emptyset$ and $f(S)=[0,1]$
$endgroup$
add a comment |
$begingroup$
This may be overkill, but the cantor function with $S=$ the middle-thirds cantor set will work:
2 properties of this function are: $S^circ = emptyset$ and $f(S)=[0,1]$
$endgroup$
This may be overkill, but the cantor function with $S=$ the middle-thirds cantor set will work:
2 properties of this function are: $S^circ = emptyset$ and $f(S)=[0,1]$
answered Dec 6 '18 at 18:00
user25959user25959
1,573816
1,573816
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