Tietze extension theorem on real line

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Let $F$ be a closed subset of $mathbb R$ and let $f:Fto mathbb R$ be a continuous function. I want to show that there exists a continuous extension of $f$ on $mathbb R$. For that I proceed as follows:

Since $mathbb Rsetminus F$ is open there exists countable collection $(a_n,b_n):nin mathbb N$ of pairwise disjoint open intervals such that $mathbb Rsetminus F=bigcuplimits_n=1^infty (a_n,b_n)$.

We define $g:mathbb Rto mathbb R$ by $g(x)=f(x)$ if $xin F$. If $xnotin F$, there must exist unique $(a,b)$ in the above collection such that $xin (a,b)$ and we define $g(x)=fracb-xb-af(a)+fracx-ab-af(b)$ if $a,bin mathbb R$. If $a,b$ are infinite, then we adjust accordingly. I could show that if $x_0in mathbb Rsetminus F$, then $g$ is continuous at $x_0$. Problem starts if $x_0in F$. If $xin F^0$, then also I can show. But if $xin partial F$, then I could not do anything. Please help!

real-analysis general-topology continuity

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asked Jul 23 at 13:39

Anupam

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Let $F$ be a closed subset of $mathbb R$ and let $f:Fto mathbb R$ be a continuous function. I want to show that there exists a continuous extension of $f$ on $mathbb R$. For that I proceed as follows:

Since $mathbb Rsetminus F$ is open there exists countable collection $(a_n,b_n):nin mathbb N$ of pairwise disjoint open intervals such that $mathbb Rsetminus F=bigcuplimits_n=1^infty (a_n,b_n)$.

We define $g:mathbb Rto mathbb R$ by $g(x)=f(x)$ if $xin F$. If $xnotin F$, there must exist unique $(a,b)$ in the above collection such that $xin (a,b)$ and we define $g(x)=fracb-xb-af(a)+fracx-ab-af(b)$ if $a,bin mathbb R$. If $a,b$ are infinite, then we adjust accordingly. I could show that if $x_0in mathbb Rsetminus F$, then $g$ is continuous at $x_0$. Problem starts if $x_0in F$. If $xin F^0$, then also I can show. But if $xin partial F$, then I could not do anything. Please help!

real-analysis general-topology continuity

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asked Jul 23 at 13:39

Anupam

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Let $F$ be a closed subset of $mathbb R$ and let $f:Fto mathbb R$ be a continuous function. I want to show that there exists a continuous extension of $f$ on $mathbb R$. For that I proceed as follows:

Since $mathbb Rsetminus F$ is open there exists countable collection $(a_n,b_n):nin mathbb N$ of pairwise disjoint open intervals such that $mathbb Rsetminus F=bigcuplimits_n=1^infty (a_n,b_n)$.

We define $g:mathbb Rto mathbb R$ by $g(x)=f(x)$ if $xin F$. If $xnotin F$, there must exist unique $(a,b)$ in the above collection such that $xin (a,b)$ and we define $g(x)=fracb-xb-af(a)+fracx-ab-af(b)$ if $a,bin mathbb R$. If $a,b$ are infinite, then we adjust accordingly. I could show that if $x_0in mathbb Rsetminus F$, then $g$ is continuous at $x_0$. Problem starts if $x_0in F$. If $xin F^0$, then also I can show. But if $xin partial F$, then I could not do anything. Please help!

real-analysis general-topology continuity

share|cite|improve this question

asked Jul 23 at 13:39

Anupam

2,1781822

Let $F$ be a closed subset of $mathbb R$ and let $f:Fto mathbb R$ be a continuous function. I want to show that there exists a continuous extension of $f$ on $mathbb R$. For that I proceed as follows:

Since $mathbb Rsetminus F$ is open there exists countable collection $(a_n,b_n):nin mathbb N$ of pairwise disjoint open intervals such that $mathbb Rsetminus F=bigcuplimits_n=1^infty (a_n,b_n)$.

We define $g:mathbb Rto mathbb R$ by $g(x)=f(x)$ if $xin F$. If $xnotin F$, there must exist unique $(a,b)$ in the above collection such that $xin (a,b)$ and we define $g(x)=fracb-xb-af(a)+fracx-ab-af(b)$ if $a,bin mathbb R$. If $a,b$ are infinite, then we adjust accordingly. I could show that if $x_0in mathbb Rsetminus F$, then $g$ is continuous at $x_0$. Problem starts if $x_0in F$. If $xin F^0$, then also I can show. But if $xin partial F$, then I could not do anything. Please help!

real-analysis general-topology continuity

share|cite|improve this question

asked Jul 23 at 13:39

Anupam

2,1781822

share|cite|improve this question

share|cite|improve this question

asked Jul 23 at 13:39

Anupam

2,1781822

asked Jul 23 at 13:39

Anupam

2,1781822

asked Jul 23 at 13:39

Anupam

2,1781822

2,1781822

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add a comment | 

2 Answers
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Hint: The solution is probably going to use this inequality: If $0le tle 1$ then $$beginalign|f(x)-(tf(a)+(1-t)f(b))|&=|t(f(x)-f(a))+(1-t)(f(x)-f(b))|
\&le|t(f(x)-f(a))|+|(1-t)(f(x)-f(b))|
\&=t|f(x)-f(a)|+(1-t)|f(x)-f(b)|
\&lemax(|f(x)-f(a)|,|f(x)-f(b)|).endalign$$

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answered Jul 23 at 14:07

David C. Ullrich

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Get $xinpartial F$. Let's show the left continuity of $g$ in $x$. It is enough to show that for a strict increasing sequence $(x_n)_ninmathbbR$ such that $x_nuparrow x$ we have that:
$$g(x_n)rightarrow f(x), nrightarrow infty.$$
Now, if $ x_nnot in F$ is finite, the conclusion follows by the continuity of $f$. If not, it is enough to show that, if $(x_n_k)_kinmathbbN$ is the subsequence of $(x_n)_ninmathbbN$ whose elements are not in $F$, then $$g(x_n_k)rightarrow f(x), krightarrow infty.$$
Now, if there exists an interval $(a,x)$ in the complement of $F$, we are done by the fact that $g$ is continuous on $(a,x)$ and $lim_yrightarrow x^-g(y)=f(x)$. If not, for each $kinmathbbN$, let $m_kinmathbbN$ be the unique integer such that $x_n_kin (a_m_k,b_m_k)$. Then, in order:

1. $forall kinmathbbN, a_m_k,b_m_kin F$;


2. $a_m_krightarrow x, krightarrowinfty$ and $b_m_k rightarrow x, krightarrow infty$;


3. $f(a_m_k)rightarrow f(x), krightarrowinfty$ and $f(b_m_k)rightarrow f(x), krightarrowinfty$;


4. $forall k in mathbbN,g(x_n_k)in[min(f(a_m_k),f(b_m_k)),max(f(a_m_k),f(b_m_k))]$


5. $g(x_n_k)rightarrow f(x), krightarrowinfty$.

By the arbitrariness of $(x_n)_ninmathbbN$, $g$ is left continuous at $x$. Analogously you can prove the right continuity of $g$ in $x$.

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edited Jul 24 at 12:34

answered Jul 23 at 14:40

Bob

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Hint: The solution is probably going to use this inequality: If $0le tle 1$ then $$beginalign|f(x)-(tf(a)+(1-t)f(b))|&=|t(f(x)-f(a))+(1-t)(f(x)-f(b))|
\&le|t(f(x)-f(a))|+|(1-t)(f(x)-f(b))|
\&=t|f(x)-f(a)|+(1-t)|f(x)-f(b)|
\&lemax(|f(x)-f(a)|,|f(x)-f(b)|).endalign$$

share|cite|improve this answer

answered Jul 23 at 14:07

David C. Ullrich

54.1k33481

add a comment | 

up vote
1
down vote

Hint: The solution is probably going to use this inequality: If $0le tle 1$ then $$beginalign|f(x)-(tf(a)+(1-t)f(b))|&=|t(f(x)-f(a))+(1-t)(f(x)-f(b))|
\&le|t(f(x)-f(a))|+|(1-t)(f(x)-f(b))|
\&=t|f(x)-f(a)|+(1-t)|f(x)-f(b)|
\&lemax(|f(x)-f(a)|,|f(x)-f(b)|).endalign$$

share|cite|improve this answer

answered Jul 23 at 14:07

David C. Ullrich

54.1k33481

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hint: The solution is probably going to use this inequality: If $0le tle 1$ then $$beginalign|f(x)-(tf(a)+(1-t)f(b))|&=|t(f(x)-f(a))+(1-t)(f(x)-f(b))|
\&le|t(f(x)-f(a))|+|(1-t)(f(x)-f(b))|
\&=t|f(x)-f(a)|+(1-t)|f(x)-f(b)|
\&lemax(|f(x)-f(a)|,|f(x)-f(b)|).endalign$$

share|cite|improve this answer

answered Jul 23 at 14:07

David C. Ullrich

54.1k33481

Hint: The solution is probably going to use this inequality: If $0le tle 1$ then $$beginalign|f(x)-(tf(a)+(1-t)f(b))|&=|t(f(x)-f(a))+(1-t)(f(x)-f(b))|
\&le|t(f(x)-f(a))|+|(1-t)(f(x)-f(b))|
\&=t|f(x)-f(a)|+(1-t)|f(x)-f(b)|
\&lemax(|f(x)-f(a)|,|f(x)-f(b)|).endalign$$

share|cite|improve this answer

answered Jul 23 at 14:07

David C. Ullrich

54.1k33481

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 14:07

David C. Ullrich

54.1k33481

answered Jul 23 at 14:07

David C. Ullrich

54.1k33481

answered Jul 23 at 14:07

David C. Ullrich

54.1k33481

54.1k33481

add a comment | 

add a comment | 

up vote
0
down vote

Get $xinpartial F$. Let's show the left continuity of $g$ in $x$. It is enough to show that for a strict increasing sequence $(x_n)_ninmathbbR$ such that $x_nuparrow x$ we have that:
$$g(x_n)rightarrow f(x), nrightarrow infty.$$
Now, if $ x_nnot in F$ is finite, the conclusion follows by the continuity of $f$. If not, it is enough to show that, if $(x_n_k)_kinmathbbN$ is the subsequence of $(x_n)_ninmathbbN$ whose elements are not in $F$, then $$g(x_n_k)rightarrow f(x), krightarrow infty.$$
Now, if there exists an interval $(a,x)$ in the complement of $F$, we are done by the fact that $g$ is continuous on $(a,x)$ and $lim_yrightarrow x^-g(y)=f(x)$. If not, for each $kinmathbbN$, let $m_kinmathbbN$ be the unique integer such that $x_n_kin (a_m_k,b_m_k)$. Then, in order:

1. $forall kinmathbbN, a_m_k,b_m_kin F$;


2. $a_m_krightarrow x, krightarrowinfty$ and $b_m_k rightarrow x, krightarrow infty$;


3. $f(a_m_k)rightarrow f(x), krightarrowinfty$ and $f(b_m_k)rightarrow f(x), krightarrowinfty$;


4. $forall k in mathbbN,g(x_n_k)in[min(f(a_m_k),f(b_m_k)),max(f(a_m_k),f(b_m_k))]$


5. $g(x_n_k)rightarrow f(x), krightarrowinfty$.

By the arbitrariness of $(x_n)_ninmathbbN$, $g$ is left continuous at $x$. Analogously you can prove the right continuity of $g$ in $x$.

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edited Jul 24 at 12:34

answered Jul 23 at 14:40

Bob

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up vote
0
down vote

Get $xinpartial F$. Let's show the left continuity of $g$ in $x$. It is enough to show that for a strict increasing sequence $(x_n)_ninmathbbR$ such that $x_nuparrow x$ we have that:
$$g(x_n)rightarrow f(x), nrightarrow infty.$$
Now, if $ x_nnot in F$ is finite, the conclusion follows by the continuity of $f$. If not, it is enough to show that, if $(x_n_k)_kinmathbbN$ is the subsequence of $(x_n)_ninmathbbN$ whose elements are not in $F$, then $$g(x_n_k)rightarrow f(x), krightarrow infty.$$
Now, if there exists an interval $(a,x)$ in the complement of $F$, we are done by the fact that $g$ is continuous on $(a,x)$ and $lim_yrightarrow x^-g(y)=f(x)$. If not, for each $kinmathbbN$, let $m_kinmathbbN$ be the unique integer such that $x_n_kin (a_m_k,b_m_k)$. Then, in order:

1. $forall kinmathbbN, a_m_k,b_m_kin F$;


2. $a_m_krightarrow x, krightarrowinfty$ and $b_m_k rightarrow x, krightarrow infty$;


3. $f(a_m_k)rightarrow f(x), krightarrowinfty$ and $f(b_m_k)rightarrow f(x), krightarrowinfty$;


4. $forall k in mathbbN,g(x_n_k)in[min(f(a_m_k),f(b_m_k)),max(f(a_m_k),f(b_m_k))]$


5. $g(x_n_k)rightarrow f(x), krightarrowinfty$.

By the arbitrariness of $(x_n)_ninmathbbN$, $g$ is left continuous at $x$. Analogously you can prove the right continuity of $g$ in $x$.

share|cite|improve this answer

edited Jul 24 at 12:34

answered Jul 23 at 14:40

Bob

1,517522

add a comment | 

up vote
0
down vote

up vote
0
down vote

Get $xinpartial F$. Let's show the left continuity of $g$ in $x$. It is enough to show that for a strict increasing sequence $(x_n)_ninmathbbR$ such that $x_nuparrow x$ we have that:
$$g(x_n)rightarrow f(x), nrightarrow infty.$$
Now, if $ x_nnot in F$ is finite, the conclusion follows by the continuity of $f$. If not, it is enough to show that, if $(x_n_k)_kinmathbbN$ is the subsequence of $(x_n)_ninmathbbN$ whose elements are not in $F$, then $$g(x_n_k)rightarrow f(x), krightarrow infty.$$
Now, if there exists an interval $(a,x)$ in the complement of $F$, we are done by the fact that $g$ is continuous on $(a,x)$ and $lim_yrightarrow x^-g(y)=f(x)$. If not, for each $kinmathbbN$, let $m_kinmathbbN$ be the unique integer such that $x_n_kin (a_m_k,b_m_k)$. Then, in order:

1. $forall kinmathbbN, a_m_k,b_m_kin F$;


2. $a_m_krightarrow x, krightarrowinfty$ and $b_m_k rightarrow x, krightarrow infty$;


3. $f(a_m_k)rightarrow f(x), krightarrowinfty$ and $f(b_m_k)rightarrow f(x), krightarrowinfty$;


4. $forall k in mathbbN,g(x_n_k)in[min(f(a_m_k),f(b_m_k)),max(f(a_m_k),f(b_m_k))]$


5. $g(x_n_k)rightarrow f(x), krightarrowinfty$.

By the arbitrariness of $(x_n)_ninmathbbN$, $g$ is left continuous at $x$. Analogously you can prove the right continuity of $g$ in $x$.

share|cite|improve this answer

edited Jul 24 at 12:34

answered Jul 23 at 14:40

Bob

1,517522

Get $xinpartial F$. Let's show the left continuity of $g$ in $x$. It is enough to show that for a strict increasing sequence $(x_n)_ninmathbbR$ such that $x_nuparrow x$ we have that:
$$g(x_n)rightarrow f(x), nrightarrow infty.$$
Now, if $ x_nnot in F$ is finite, the conclusion follows by the continuity of $f$. If not, it is enough to show that, if $(x_n_k)_kinmathbbN$ is the subsequence of $(x_n)_ninmathbbN$ whose elements are not in $F$, then $$g(x_n_k)rightarrow f(x), krightarrow infty.$$
Now, if there exists an interval $(a,x)$ in the complement of $F$, we are done by the fact that $g$ is continuous on $(a,x)$ and $lim_yrightarrow x^-g(y)=f(x)$. If not, for each $kinmathbbN$, let $m_kinmathbbN$ be the unique integer such that $x_n_kin (a_m_k,b_m_k)$. Then, in order:

1. $forall kinmathbbN, a_m_k,b_m_kin F$;


2. $a_m_krightarrow x, krightarrowinfty$ and $b_m_k rightarrow x, krightarrow infty$;


3. $f(a_m_k)rightarrow f(x), krightarrowinfty$ and $f(b_m_k)rightarrow f(x), krightarrowinfty$;


4. $forall k in mathbbN,g(x_n_k)in[min(f(a_m_k),f(b_m_k)),max(f(a_m_k),f(b_m_k))]$


5. $g(x_n_k)rightarrow f(x), krightarrowinfty$.

By the arbitrariness of $(x_n)_ninmathbbN$, $g$ is left continuous at $x$. Analogously you can prove the right continuity of $g$ in $x$.

share|cite|improve this answer

edited Jul 24 at 12:34

answered Jul 23 at 14:40

Bob

1,517522

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 12:34

edited Jul 24 at 12:34

edited Jul 24 at 12:34

answered Jul 23 at 14:40

Bob

1,517522

answered Jul 23 at 14:40

Bob

1,517522

answered Jul 23 at 14:40

Bob

1,517522

1,517522

add a comment | 

add a comment | 

 

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