Resolving a “paradox” with real continuous function The 2019 Stack Overflow Developer Survey Results Are InOscillation of a continuous $f$Convergence of a Continuous Function and Compactness of Upper Contour SetsNowhere integrable functionIs my own proof of the Bolzano-Weierstrass Theorem correct?Given $f$ continuous and $epsilon>0$, prove that exists a polygonal function $g$ such that $|f(x)-g(x)|<epsilon$ for every $x in [a,b]$A question about Existence of a Continuous function.Continuous function in compact sets - Real analysisProb. 14, Sec. 5.4, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: Every continuous periodic function is bounded and uniformly continuousUnderstanding this statement for the properties of continuous functionsContinuous Functions and Integrability
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Resolving a “paradox” with real continuous function
The 2019 Stack Overflow Developer Survey Results Are InOscillation of a continuous $f$Convergence of a Continuous Function and Compactness of Upper Contour SetsNowhere integrable functionIs my own proof of the Bolzano-Weierstrass Theorem correct?Given $f$ continuous and $epsilon>0$, prove that exists a polygonal function $g$ such that $|f(x)-g(x)|<epsilon$ for every $x in [a,b]$A question about Existence of a Continuous function.Continuous function in compact sets - Real analysisProb. 14, Sec. 5.4, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: Every continuous periodic function is bounded and uniformly continuousUnderstanding this statement for the properties of continuous functionsContinuous Functions and Integrability
$begingroup$
Let
$$fin mathcal C^0(mathbb R),quad f(0)=0,quad f(a)>0,quad f(-a)<0,quad a>0.$$
Let interval $I$ satisfies:
$$xin I =K,$$ where $K$ is a subinterval, $Ksubset I,$ and $0in K$.
Question: Does such an $I$ always exist for any $f$?
Yes, of course. Let $A=f(x)neq 0$. $A$ must be nonempty. Take its element, $a^*$, to be the closest element to $0$.
Similarly, let $A'=f(x)neq0$. $A'$ must be nonempty. Take its element, $a'^*$, to be the closest element to $0$. $(a'^*,a^*)$ is the interval we want.
No. Intuitively, it is easy to construct Weierstrass function $f$ such that $forall varepsilon>0$, we have $xin (0,varepsilon)$ is infinite and discrete.
I think $(2)$ is correct (so $(1)$ is wrong), because we can construct a $f$ such that $A$ is a discrete set without minimum, and $inf (A)=0$. Does this make sense?
A related question: Does there exist an interval with singleton interior (here "interior" means topological interior)?
real-analysis calculus analysis functions continuity
edited Mar 23 at 5:50
MarianD
2,2611618
asked Mar 23 at 3:51
High GPAHigh GPA
916422
$endgroup$
|
show 3 more comments
$begingroup$
Let
$$fin mathcal C^0(mathbb R),quad f(0)=0,quad f(a)>0,quad f(-a)<0,quad a>0.$$
Let interval $I$ satisfies:
$$xin I =K,$$ where $K$ is a subinterval, $Ksubset I,$ and $0in K$.
Question: Does such an $I$ always exist for any $f$?
Yes, of course. Let $A=f(x)neq 0$. $A$ must be nonempty. Take its element, $a^*$, to be the closest element to $0$.
Similarly, let $A'=f(x)neq0$. $A'$ must be nonempty. Take its element, $a'^*$, to be the closest element to $0$. $(a'^*,a^*)$ is the interval we want.
No. Intuitively, it is easy to construct Weierstrass function $f$ such that $forall varepsilon>0$, we have $xin (0,varepsilon)$ is infinite and discrete.
I think $(2)$ is correct (so $(1)$ is wrong), because we can construct a $f$ such that $A$ is a discrete set without minimum, and $inf (A)=0$. Does this make sense?
A related question: Does there exist an interval with singleton interior (here "interior" means topological interior)?
real-analysis calculus analysis functions continuity
edited Mar 23 at 5:50
MarianD
2,2611618
asked Mar 23 at 3:51
High GPAHigh GPA
916422
$endgroup$
1
$begingroup$
1 is wrong because you have no guarantee that $f$ is actually zero on every point of the interval. If you have a potential example for $2$, then try running the argument in (1) with that specific function to see the error. Second question: depends on whether you consider $[a,a]=a$ an interval, and whether if you do, you consider $a$ to be in the interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:06
$begingroup$
In regards to the latter question, an interval with a point as interior has that point alone as boundary, and contains that point, obviously, so a singleton is always a closed set. As to whether to call it an interval, well, I dunno. Intervals are usually meant to indicate the presence of uncountably many points on the line. A single point is obviously countable.
$endgroup$
– Allawonder
Mar 23 at 4:21
$begingroup$
@ArturoMagidin Thank you! I though $int(S)=Ssetminus bd(S)$, so $int([a,a])$ is empty?
$endgroup$
– High GPA
Mar 23 at 4:42
$begingroup$
Like I said: depends on your definitions. Some people consider $[a,a]$ an interval, some do not.
$endgroup$
– Arturo Magidin
Mar 23 at 4:43
2
$begingroup$
Topological interior is the largest open set contained in the set; if that’s what you meant, then the degenerate interval has empty interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:46
|
show 3 more comments
$begingroup$
Let
$$fin mathcal C^0(mathbb R),quad f(0)=0,quad f(a)>0,quad f(-a)<0,quad a>0.$$
Let interval $I$ satisfies:
$$xin I =K,$$ where $K$ is a subinterval, $Ksubset I,$ and $0in K$.
Question: Does such an $I$ always exist for any $f$?
Yes, of course. Let $A=f(x)neq 0$. $A$ must be nonempty. Take its element, $a^*$, to be the closest element to $0$.
Similarly, let $A'=f(x)neq0$. $A'$ must be nonempty. Take its element, $a'^*$, to be the closest element to $0$. $(a'^*,a^*)$ is the interval we want.
No. Intuitively, it is easy to construct Weierstrass function $f$ such that $forall varepsilon>0$, we have $xin (0,varepsilon)$ is infinite and discrete.
I think $(2)$ is correct (so $(1)$ is wrong), because we can construct a $f$ such that $A$ is a discrete set without minimum, and $inf (A)=0$. Does this make sense?
A related question: Does there exist an interval with singleton interior (here "interior" means topological interior)?
real-analysis calculus analysis functions continuity
edited Mar 23 at 5:50
MarianD
2,2611618
asked Mar 23 at 3:51
High GPAHigh GPA
916422
$endgroup$
Let
$$fin mathcal C^0(mathbb R),quad f(0)=0,quad f(a)>0,quad f(-a)<0,quad a>0.$$
Let interval $I$ satisfies:
$$xin I =K,$$ where $K$ is a subinterval, $Ksubset I,$ and $0in K$.
Question: Does such an $I$ always exist for any $f$?
Yes, of course. Let $A=f(x)neq 0$. $A$ must be nonempty. Take its element, $a^*$, to be the closest element to $0$.
Similarly, let $A'=f(x)neq0$. $A'$ must be nonempty. Take its element, $a'^*$, to be the closest element to $0$. $(a'^*,a^*)$ is the interval we want.
No. Intuitively, it is easy to construct Weierstrass function $f$ such that $forall varepsilon>0$, we have $xin (0,varepsilon)$ is infinite and discrete.
I think $(2)$ is correct (so $(1)$ is wrong), because we can construct a $f$ such that $A$ is a discrete set without minimum, and $inf (A)=0$. Does this make sense?
A related question: Does there exist an interval with singleton interior (here "interior" means topological interior)?
real-analysis calculus analysis functions continuity
real-analysis calculus analysis functions continuity
edited Mar 23 at 5:50
MarianD
2,2611618
asked Mar 23 at 3:51
High GPAHigh GPA
916422
edited Mar 23 at 5:50
MarianD
2,2611618
asked Mar 23 at 3:51
High GPAHigh GPA
916422
edited Mar 23 at 5:50
MarianD
2,2611618
edited Mar 23 at 5:50
MarianD
2,2611618
edited Mar 23 at 5:50
MarianD
2,2611618
2,2611618
asked Mar 23 at 3:51
High GPAHigh GPA
916422
asked Mar 23 at 3:51
High GPAHigh GPA
916422
asked Mar 23 at 3:51
High GPAHigh GPA
916422
916422
1
$begingroup$
1 is wrong because you have no guarantee that $f$ is actually zero on every point of the interval. If you have a potential example for $2$, then try running the argument in (1) with that specific function to see the error. Second question: depends on whether you consider $[a,a]=a$ an interval, and whether if you do, you consider $a$ to be in the interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:06
$begingroup$
In regards to the latter question, an interval with a point as interior has that point alone as boundary, and contains that point, obviously, so a singleton is always a closed set. As to whether to call it an interval, well, I dunno. Intervals are usually meant to indicate the presence of uncountably many points on the line. A single point is obviously countable.
$endgroup$
– Allawonder
Mar 23 at 4:21
$begingroup$
@ArturoMagidin Thank you! I though $int(S)=Ssetminus bd(S)$, so $int([a,a])$ is empty?
$endgroup$
– High GPA
Mar 23 at 4:42
$begingroup$
Like I said: depends on your definitions. Some people consider $[a,a]$ an interval, some do not.
$endgroup$
– Arturo Magidin
Mar 23 at 4:43
2
$begingroup$
Topological interior is the largest open set contained in the set; if that’s what you meant, then the degenerate interval has empty interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:46
|
show 3 more comments
1
$begingroup$
1 is wrong because you have no guarantee that $f$ is actually zero on every point of the interval. If you have a potential example for $2$, then try running the argument in (1) with that specific function to see the error. Second question: depends on whether you consider $[a,a]=a$ an interval, and whether if you do, you consider $a$ to be in the interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:06
$begingroup$
In regards to the latter question, an interval with a point as interior has that point alone as boundary, and contains that point, obviously, so a singleton is always a closed set. As to whether to call it an interval, well, I dunno. Intervals are usually meant to indicate the presence of uncountably many points on the line. A single point is obviously countable.
$endgroup$
– Allawonder
Mar 23 at 4:21
$begingroup$
@ArturoMagidin Thank you! I though $int(S)=Ssetminus bd(S)$, so $int([a,a])$ is empty?
$endgroup$
– High GPA
Mar 23 at 4:42
$begingroup$
Like I said: depends on your definitions. Some people consider $[a,a]$ an interval, some do not.
$endgroup$
– Arturo Magidin
Mar 23 at 4:43
2
$begingroup$
Topological interior is the largest open set contained in the set; if that’s what you meant, then the degenerate interval has empty interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:46
1
1
$begingroup$
1 is wrong because you have no guarantee that $f$ is actually zero on every point of the interval. If you have a potential example for $2$, then try running the argument in (1) with that specific function to see the error. Second question: depends on whether you consider $[a,a]=a$ an interval, and whether if you do, you consider $a$ to be in the interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:06
$begingroup$
1 is wrong because you have no guarantee that $f$ is actually zero on every point of the interval. If you have a potential example for $2$, then try running the argument in (1) with that specific function to see the error. Second question: depends on whether you consider $[a,a]=a$ an interval, and whether if you do, you consider $a$ to be in the interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:06
$begingroup$
In regards to the latter question, an interval with a point as interior has that point alone as boundary, and contains that point, obviously, so a singleton is always a closed set. As to whether to call it an interval, well, I dunno. Intervals are usually meant to indicate the presence of uncountably many points on the line. A single point is obviously countable.
$endgroup$
– Allawonder
Mar 23 at 4:21
$begingroup$
In regards to the latter question, an interval with a point as interior has that point alone as boundary, and contains that point, obviously, so a singleton is always a closed set. As to whether to call it an interval, well, I dunno. Intervals are usually meant to indicate the presence of uncountably many points on the line. A single point is obviously countable.
$endgroup$
– Allawonder
Mar 23 at 4:21
$begingroup$
@ArturoMagidin Thank you! I though $int(S)=Ssetminus bd(S)$, so $int([a,a])$ is empty?
$endgroup$
– High GPA
Mar 23 at 4:42
$begingroup$
@ArturoMagidin Thank you! I though $int(S)=Ssetminus bd(S)$, so $int([a,a])$ is empty?
$endgroup$
– High GPA
Mar 23 at 4:42
$begingroup$
Like I said: depends on your definitions. Some people consider $[a,a]$ an interval, some do not.
$endgroup$
– Arturo Magidin
Mar 23 at 4:43
$begingroup$
Like I said: depends on your definitions. Some people consider $[a,a]$ an interval, some do not.
$endgroup$
– Arturo Magidin
Mar 23 at 4:43
2
2
$begingroup$
Topological interior is the largest open set contained in the set; if that’s what you meant, then the degenerate interval has empty interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:46
$begingroup$
Topological interior is the largest open set contained in the set; if that’s what you meant, then the degenerate interval has empty interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:46
|
show 3 more comments
0
active
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1
$begingroup$
1 is wrong because you have no guarantee that $f$ is actually zero on every point of the interval. If you have a potential example for $2$, then try running the argument in (1) with that specific function to see the error. Second question: depends on whether you consider $[a,a]=a$ an interval, and whether if you do, you consider $a$ to be in the interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:06
$begingroup$
In regards to the latter question, an interval with a point as interior has that point alone as boundary, and contains that point, obviously, so a singleton is always a closed set. As to whether to call it an interval, well, I dunno. Intervals are usually meant to indicate the presence of uncountably many points on the line. A single point is obviously countable.
$endgroup$
– Allawonder
Mar 23 at 4:21
$begingroup$
@ArturoMagidin Thank you! I though $int(S)=Ssetminus bd(S)$, so $int([a,a])$ is empty?
$endgroup$
– High GPA
Mar 23 at 4:42
$begingroup$
Like I said: depends on your definitions. Some people consider $[a,a]$ an interval, some do not.
$endgroup$
– Arturo Magidin
Mar 23 at 4:43
2
$begingroup$
Topological interior is the largest open set contained in the set; if that’s what you meant, then the degenerate interval has empty interior.
$endgroup$
– Arturo Magidin
Mar 23 at 4:46