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When do we need to show a function is well-defined?
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Proposition: Let $A$ be a dense subset of $B$. Let $f$ be a uniformly continuous function from $A$ to a complete metric space $C$. Then $f$ has a unique extension to a continuous function from $B$ to $C$.
My attempt, only on showing $f$ has an extension to a continuous function from $B$ to $C$:
Since $(A,d)subseteq (B,d)$ such that $overlineA=B$, then $forall bin B$, given any $epsilon gt 0$, there exists $ain A$ such that $a in S_epsilon(b)$. Due to Theorem 36, there exists a sequence $alpha_n$ that converges to $b$. By Theorem 43, $alpha_n$ is a Cauchy sequence. By Theorem 51, since $f$ is continuous, $f(a_n)$ is also a Cauchy sequence in $C$. Since $C$ is complete, $f(alpha_n)$ converges to some $gamma in C$; Hence, $f(b)=c$.
But, I was told that this attempt at a proof is insufficient to prove the claim since I still need to show that $f$ is well-defined; which is also what the proof from Kaplansky's text has done.
Question 1: I do not understand why proving $f$ is well-defined is necessary due to my understanding that since $a_n$ arbitrary in the set $S$ of sequences that converges to $b$, then we no longer need to consider another sequence in $S$. But apparently, I am wrong. Why?
Question 2: Why is it allowed to assumed that the sequence at (16), seen below has a subsequence converging to $c$?
functions convergence metric-spaces continuity cauchy-sequences
asked Jul 30 at 6:06
TheLast Cipher
538414
add a comment |Â
up vote
0
down vote
favorite
Proposition: Let $A$ be a dense subset of $B$. Let $f$ be a uniformly continuous function from $A$ to a complete metric space $C$. Then $f$ has a unique extension to a continuous function from $B$ to $C$.
My attempt, only on showing $f$ has an extension to a continuous function from $B$ to $C$:
Since $(A,d)subseteq (B,d)$ such that $overlineA=B$, then $forall bin B$, given any $epsilon gt 0$, there exists $ain A$ such that $a in S_epsilon(b)$. Due to Theorem 36, there exists a sequence $alpha_n$ that converges to $b$. By Theorem 43, $alpha_n$ is a Cauchy sequence. By Theorem 51, since $f$ is continuous, $f(a_n)$ is also a Cauchy sequence in $C$. Since $C$ is complete, $f(alpha_n)$ converges to some $gamma in C$; Hence, $f(b)=c$.
But, I was told that this attempt at a proof is insufficient to prove the claim since I still need to show that $f$ is well-defined; which is also what the proof from Kaplansky's text has done.
Question 1: I do not understand why proving $f$ is well-defined is necessary due to my understanding that since $a_n$ arbitrary in the set $S$ of sequences that converges to $b$, then we no longer need to consider another sequence in $S$. But apparently, I am wrong. Why?
Question 2: Why is it allowed to assumed that the sequence at (16), seen below has a subsequence converging to $c$?
functions convergence metric-spaces continuity cauchy-sequences
asked Jul 30 at 6:06
TheLast Cipher
538414
If you don't show that $f$ is well defined you won't be able to show that the function you have defined on $B$ is continuous. The answer to your second quesion is if you pick the 1st, 3rd, 5th ... elements in the sequence you get a desired subsequence. As far as your comment below is concerned, you will have to write down the proof of continuity to convince yourself that you have a problem.
– Kavi Rama Murthy
Jul 30 at 6:11
Why? Is $a_n$ not arbitrary?
– TheLast Cipher
Jul 30 at 6:12
I know picking elements from the sequence $f(x_i)$ would get me a subsequence. My question was why can we assume that its subsequence converges to $c$. To emphasize, my question was regarding its convergence to $c$. Thanks.
– TheLast Cipher
Jul 30 at 6:21
See the definition of $c$.
– Kavi Rama Murthy
Jul 30 at 6:27
add a comment |Â
up vote
0
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favorite
up vote
0
down vote
favorite
Proposition: Let $A$ be a dense subset of $B$. Let $f$ be a uniformly continuous function from $A$ to a complete metric space $C$. Then $f$ has a unique extension to a continuous function from $B$ to $C$.
My attempt, only on showing $f$ has an extension to a continuous function from $B$ to $C$:
Since $(A,d)subseteq (B,d)$ such that $overlineA=B$, then $forall bin B$, given any $epsilon gt 0$, there exists $ain A$ such that $a in S_epsilon(b)$. Due to Theorem 36, there exists a sequence $alpha_n$ that converges to $b$. By Theorem 43, $alpha_n$ is a Cauchy sequence. By Theorem 51, since $f$ is continuous, $f(a_n)$ is also a Cauchy sequence in $C$. Since $C$ is complete, $f(alpha_n)$ converges to some $gamma in C$; Hence, $f(b)=c$.
But, I was told that this attempt at a proof is insufficient to prove the claim since I still need to show that $f$ is well-defined; which is also what the proof from Kaplansky's text has done.
Question 1: I do not understand why proving $f$ is well-defined is necessary due to my understanding that since $a_n$ arbitrary in the set $S$ of sequences that converges to $b$, then we no longer need to consider another sequence in $S$. But apparently, I am wrong. Why?
Question 2: Why is it allowed to assumed that the sequence at (16), seen below has a subsequence converging to $c$?
functions convergence metric-spaces continuity cauchy-sequences
asked Jul 30 at 6:06
TheLast Cipher
538414
Proposition: Let $A$ be a dense subset of $B$. Let $f$ be a uniformly continuous function from $A$ to a complete metric space $C$. Then $f$ has a unique extension to a continuous function from $B$ to $C$.
My attempt, only on showing $f$ has an extension to a continuous function from $B$ to $C$:
Since $(A,d)subseteq (B,d)$ such that $overlineA=B$, then $forall bin B$, given any $epsilon gt 0$, there exists $ain A$ such that $a in S_epsilon(b)$. Due to Theorem 36, there exists a sequence $alpha_n$ that converges to $b$. By Theorem 43, $alpha_n$ is a Cauchy sequence. By Theorem 51, since $f$ is continuous, $f(a_n)$ is also a Cauchy sequence in $C$. Since $C$ is complete, $f(alpha_n)$ converges to some $gamma in C$; Hence, $f(b)=c$.
But, I was told that this attempt at a proof is insufficient to prove the claim since I still need to show that $f$ is well-defined; which is also what the proof from Kaplansky's text has done.
Question 1: I do not understand why proving $f$ is well-defined is necessary due to my understanding that since $a_n$ arbitrary in the set $S$ of sequences that converges to $b$, then we no longer need to consider another sequence in $S$. But apparently, I am wrong. Why?
Question 2: Why is it allowed to assumed that the sequence at (16), seen below has a subsequence converging to $c$?
functions convergence metric-spaces continuity cauchy-sequences
asked Jul 30 at 6:06
TheLast Cipher
538414
asked Jul 30 at 6:06
TheLast Cipher
538414
asked Jul 30 at 6:06
TheLast Cipher
538414
asked Jul 30 at 6:06
TheLast Cipher
538414
538414
If you don't show that $f$ is well defined you won't be able to show that the function you have defined on $B$ is continuous. The answer to your second quesion is if you pick the 1st, 3rd, 5th ... elements in the sequence you get a desired subsequence. As far as your comment below is concerned, you will have to write down the proof of continuity to convince yourself that you have a problem.
– Kavi Rama Murthy
Jul 30 at 6:11
Why? Is $a_n$ not arbitrary?
– TheLast Cipher
Jul 30 at 6:12
I know picking elements from the sequence $f(x_i)$ would get me a subsequence. My question was why can we assume that its subsequence converges to $c$. To emphasize, my question was regarding its convergence to $c$. Thanks.
– TheLast Cipher
Jul 30 at 6:21
See the definition of $c$.
– Kavi Rama Murthy
Jul 30 at 6:27
add a comment |Â
If you don't show that $f$ is well defined you won't be able to show that the function you have defined on $B$ is continuous. The answer to your second quesion is if you pick the 1st, 3rd, 5th ... elements in the sequence you get a desired subsequence. As far as your comment below is concerned, you will have to write down the proof of continuity to convince yourself that you have a problem.
– Kavi Rama Murthy
Jul 30 at 6:11
Why? Is $a_n$ not arbitrary?
– TheLast Cipher
Jul 30 at 6:12
I know picking elements from the sequence $f(x_i)$ would get me a subsequence. My question was why can we assume that its subsequence converges to $c$. To emphasize, my question was regarding its convergence to $c$. Thanks.
– TheLast Cipher
Jul 30 at 6:21
See the definition of $c$.
– Kavi Rama Murthy
Jul 30 at 6:27
If you don't show that $f$ is well defined you won't be able to show that the function you have defined on $B$ is continuous. The answer to your second quesion is if you pick the 1st, 3rd, 5th ... elements in the sequence you get a desired subsequence. As far as your comment below is concerned, you will have to write down the proof of continuity to convince yourself that you have a problem.
– Kavi Rama Murthy
Jul 30 at 6:11
If you don't show that $f$ is well defined you won't be able to show that the function you have defined on $B$ is continuous. The answer to your second quesion is if you pick the 1st, 3rd, 5th ... elements in the sequence you get a desired subsequence. As far as your comment below is concerned, you will have to write down the proof of continuity to convince yourself that you have a problem.
– Kavi Rama Murthy
Jul 30 at 6:11
Why? Is $a_n$ not arbitrary?
– TheLast Cipher
Jul 30 at 6:12
Why? Is $a_n$ not arbitrary?
– TheLast Cipher
Jul 30 at 6:12
I know picking elements from the sequence $f(x_i)$ would get me a subsequence. My question was why can we assume that its subsequence converges to $c$. To emphasize, my question was regarding its convergence to $c$. Thanks.
– TheLast Cipher
Jul 30 at 6:21
I know picking elements from the sequence $f(x_i)$ would get me a subsequence. My question was why can we assume that its subsequence converges to $c$. To emphasize, my question was regarding its convergence to $c$. Thanks.
– TheLast Cipher
Jul 30 at 6:21
See the definition of $c$.
– Kavi Rama Murthy
Jul 30 at 6:27
See the definition of $c$.
– Kavi Rama Murthy
Jul 30 at 6:27
add a comment |Â
3 Answers
3
active
oldest
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up vote
1
down vote
accepted
OK, here is my comment as an answer:
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$
that you made. You know that for any arbitrary $a_nto a$,$(f(a_n))$
converges; you do not yet know that all of the sequences $(f(a_n))$ converge to the same limit.
answered Jul 30 at 7:07
Kusma
1,027111
Just to clarify, at the end you meant "we do not yet know all sequences $(f(a_n))$ converges to the same limit."?
– TheLast Cipher
Jul 30 at 7:11
Yes, I have edited the answer to clarify.
– Kusma
Jul 30 at 7:40
Isn't this a proof of uniqueness? The question was about the extension being well-defined. That's not the same thing.
– md2perpe
Jul 30 at 18:14
add a comment |Â
up vote
0
down vote
Q1: If $(a_n)$ and $( alpha_n)$ are sequences whic both converge to $b$, then we know that the sequences $(f(a_n))$ and $f(alpha_n))$ are convergent. In order to show that $f(b)$ is well-defined, we have to show that
$$lim f(a_n)= lim f(alpha_n).$$
Q2: $(f(a_n))$ is a subsequence of the sequence in $(16)$ and $lim f(a_n)=c$.
answered Jul 30 at 6:29
Fred
37k1237
1
I know how to show that $f$ is well-defined. I just want to know why. Thanks!
– TheLast Cipher
Jul 30 at 6:30
1
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$ that you made. [You know that for any arbitrary $a_nto a$, $f(a_n)$ converges; you do not yet know that all of these sequences converge to the same limit].
– Kusma
Jul 30 at 6:42
@Kusma: I will accept your explanation if you post it as an answer. Thanks.
– TheLast Cipher
Jul 30 at 7:04
add a comment |Â
up vote
0
down vote
I don't see why it should be necessary to show that our constructed extension is well-defined, at least not because of different choices of sequences.
Define one extension $bar f : A to C$ by the following procedure:
For every $a in A$ set $bar f(a) = f(a).$ For every $b in B setminus A$ take one sequence $(a_k) subset A$ converging to $b$ (such a sequence exists by denseness of $A$) and set $bar f(b) = lim f(a_k).$ Here we need to show that the limit exists (this you might call well-definedness), but right now we don't need to show that it does not depend on the choice of sequence.
Now we have an extension. We have two steps left: 1) to show that $bar f$ is continuous, 2) to show that it is unique. One way to show the latter might be to show that it doesn't depend on the choices of sequences.
Am I wrong? Do I miss something?
answered Jul 31 at 18:33
md2perpe
5,7821922
I think you're right. I just happened to ask a question relating to the continuity of the extension of $f$ which of course requires that $overlinef$ is well-defined at $bin Bsetminus A$.
– TheLast Cipher
Aug 1 at 2:47
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
OK, here is my comment as an answer:
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$
that you made. You know that for any arbitrary $a_nto a$,$(f(a_n))$
converges; you do not yet know that all of the sequences $(f(a_n))$ converge to the same limit.
answered Jul 30 at 7:07
Kusma
1,027111
Just to clarify, at the end you meant "we do not yet know all sequences $(f(a_n))$ converges to the same limit."?
– TheLast Cipher
Jul 30 at 7:11
Yes, I have edited the answer to clarify.
– Kusma
Jul 30 at 7:40
Isn't this a proof of uniqueness? The question was about the extension being well-defined. That's not the same thing.
– md2perpe
Jul 30 at 18:14
add a comment |Â
up vote
1
down vote
accepted
OK, here is my comment as an answer:
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$
that you made. You know that for any arbitrary $a_nto a$,$(f(a_n))$
converges; you do not yet know that all of the sequences $(f(a_n))$ converge to the same limit.
answered Jul 30 at 7:07
Kusma
1,027111
Just to clarify, at the end you meant "we do not yet know all sequences $(f(a_n))$ converges to the same limit."?
– TheLast Cipher
Jul 30 at 7:11
Yes, I have edited the answer to clarify.
– Kusma
Jul 30 at 7:40
Isn't this a proof of uniqueness? The question was about the extension being well-defined. That's not the same thing.
– md2perpe
Jul 30 at 18:14
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
OK, here is my comment as an answer:
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$
that you made. You know that for any arbitrary $a_nto a$,$(f(a_n))$
converges; you do not yet know that all of the sequences $(f(a_n))$ converge to the same limit.
answered Jul 30 at 7:07
Kusma
1,027111
OK, here is my comment as an answer:
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$
that you made. You know that for any arbitrary $a_nto a$,$(f(a_n))$
converges; you do not yet know that all of the sequences $(f(a_n))$ converge to the same limit.
answered Jul 30 at 7:07
Kusma
1,027111
edited Jul 30 at 7:40
answered Jul 30 at 7:07
Kusma
1,027111
answered Jul 30 at 7:07
Kusma
1,027111
answered Jul 30 at 7:07
Kusma
1,027111
1,027111
Just to clarify, at the end you meant "we do not yet know all sequences $(f(a_n))$ converges to the same limit."?
– TheLast Cipher
Jul 30 at 7:11
Yes, I have edited the answer to clarify.
– Kusma
Jul 30 at 7:40
Isn't this a proof of uniqueness? The question was about the extension being well-defined. That's not the same thing.
– md2perpe
Jul 30 at 18:14
add a comment |Â
Just to clarify, at the end you meant "we do not yet know all sequences $(f(a_n))$ converges to the same limit."?
– TheLast Cipher
Jul 30 at 7:11
Yes, I have edited the answer to clarify.
– Kusma
Jul 30 at 7:40
Isn't this a proof of uniqueness? The question was about the extension being well-defined. That's not the same thing.
– md2perpe
Jul 30 at 18:14
Just to clarify, at the end you meant "we do not yet know all sequences $(f(a_n))$ converges to the same limit."?
– TheLast Cipher
Jul 30 at 7:11
Just to clarify, at the end you meant "we do not yet know all sequences $(f(a_n))$ converges to the same limit."?
– TheLast Cipher
Jul 30 at 7:11
Yes, I have edited the answer to clarify.
– Kusma
Jul 30 at 7:40
Yes, I have edited the answer to clarify.
– Kusma
Jul 30 at 7:40
Isn't this a proof of uniqueness? The question was about the extension being well-defined. That's not the same thing.
– md2perpe
Jul 30 at 18:14
Isn't this a proof of uniqueness? The question was about the extension being well-defined. That's not the same thing.
– md2perpe
Jul 30 at 18:14
add a comment |Â
up vote
0
down vote
Q1: If $(a_n)$ and $( alpha_n)$ are sequences whic both converge to $b$, then we know that the sequences $(f(a_n))$ and $f(alpha_n))$ are convergent. In order to show that $f(b)$ is well-defined, we have to show that
$$lim f(a_n)= lim f(alpha_n).$$
Q2: $(f(a_n))$ is a subsequence of the sequence in $(16)$ and $lim f(a_n)=c$.
answered Jul 30 at 6:29
Fred
37k1237
1
I know how to show that $f$ is well-defined. I just want to know why. Thanks!
– TheLast Cipher
Jul 30 at 6:30
1
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$ that you made. [You know that for any arbitrary $a_nto a$, $f(a_n)$ converges; you do not yet know that all of these sequences converge to the same limit].
– Kusma
Jul 30 at 6:42
@Kusma: I will accept your explanation if you post it as an answer. Thanks.
– TheLast Cipher
Jul 30 at 7:04
add a comment |Â
up vote
0
down vote
Q1: If $(a_n)$ and $( alpha_n)$ are sequences whic both converge to $b$, then we know that the sequences $(f(a_n))$ and $f(alpha_n))$ are convergent. In order to show that $f(b)$ is well-defined, we have to show that
$$lim f(a_n)= lim f(alpha_n).$$
Q2: $(f(a_n))$ is a subsequence of the sequence in $(16)$ and $lim f(a_n)=c$.
answered Jul 30 at 6:29
Fred
37k1237
1
I know how to show that $f$ is well-defined. I just want to know why. Thanks!
– TheLast Cipher
Jul 30 at 6:30
1
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$ that you made. [You know that for any arbitrary $a_nto a$, $f(a_n)$ converges; you do not yet know that all of these sequences converge to the same limit].
– Kusma
Jul 30 at 6:42
@Kusma: I will accept your explanation if you post it as an answer. Thanks.
– TheLast Cipher
Jul 30 at 7:04
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Q1: If $(a_n)$ and $( alpha_n)$ are sequences whic both converge to $b$, then we know that the sequences $(f(a_n))$ and $f(alpha_n))$ are convergent. In order to show that $f(b)$ is well-defined, we have to show that
$$lim f(a_n)= lim f(alpha_n).$$
Q2: $(f(a_n))$ is a subsequence of the sequence in $(16)$ and $lim f(a_n)=c$.
answered Jul 30 at 6:29
Fred
37k1237
Q1: If $(a_n)$ and $( alpha_n)$ are sequences whic both converge to $b$, then we know that the sequences $(f(a_n))$ and $f(alpha_n))$ are convergent. In order to show that $f(b)$ is well-defined, we have to show that
$$lim f(a_n)= lim f(alpha_n).$$
Q2: $(f(a_n))$ is a subsequence of the sequence in $(16)$ and $lim f(a_n)=c$.
answered Jul 30 at 6:29
Fred
37k1237
answered Jul 30 at 6:29
Fred
37k1237
answered Jul 30 at 6:29
Fred
37k1237
answered Jul 30 at 6:29
Fred
37k1237
37k1237
1
I know how to show that $f$ is well-defined. I just want to know why. Thanks!
– TheLast Cipher
Jul 30 at 6:30
1
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$ that you made. [You know that for any arbitrary $a_nto a$, $f(a_n)$ converges; you do not yet know that all of these sequences converge to the same limit].
– Kusma
Jul 30 at 6:42
@Kusma: I will accept your explanation if you post it as an answer. Thanks.
– TheLast Cipher
Jul 30 at 7:04
add a comment |Â
1
I know how to show that $f$ is well-defined. I just want to know why. Thanks!
– TheLast Cipher
Jul 30 at 6:30
1
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$ that you made. [You know that for any arbitrary $a_nto a$, $f(a_n)$ converges; you do not yet know that all of these sequences converge to the same limit].
– Kusma
Jul 30 at 6:42
@Kusma: I will accept your explanation if you post it as an answer. Thanks.
– TheLast Cipher
Jul 30 at 7:04
1
1
I know how to show that $f$ is well-defined. I just want to know why. Thanks!
– TheLast Cipher
Jul 30 at 6:30
I know how to show that $f$ is well-defined. I just want to know why. Thanks!
– TheLast Cipher
Jul 30 at 6:30
1
1
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$ that you made. [You know that for any arbitrary $a_nto a$, $f(a_n)$ converges; you do not yet know that all of these sequences converge to the same limit].
– Kusma
Jul 30 at 6:42
When you set $f(b)=lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(alpha_n)$ with $alpha_nto b$ and would like to set $f(b)=lim f(alpha_n)$. You both have the same definition only if $lim f(a_n)=lim f(alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$ that you made. [You know that for any arbitrary $a_nto a$, $f(a_n)$ converges; you do not yet know that all of these sequences converge to the same limit].
– Kusma
Jul 30 at 6:42
@Kusma: I will accept your explanation if you post it as an answer. Thanks.
– TheLast Cipher
Jul 30 at 7:04
@Kusma: I will accept your explanation if you post it as an answer. Thanks.
– TheLast Cipher
Jul 30 at 7:04
add a comment |Â
up vote
0
down vote
I don't see why it should be necessary to show that our constructed extension is well-defined, at least not because of different choices of sequences.
Define one extension $bar f : A to C$ by the following procedure:
For every $a in A$ set $bar f(a) = f(a).$ For every $b in B setminus A$ take one sequence $(a_k) subset A$ converging to $b$ (such a sequence exists by denseness of $A$) and set $bar f(b) = lim f(a_k).$ Here we need to show that the limit exists (this you might call well-definedness), but right now we don't need to show that it does not depend on the choice of sequence.
Now we have an extension. We have two steps left: 1) to show that $bar f$ is continuous, 2) to show that it is unique. One way to show the latter might be to show that it doesn't depend on the choices of sequences.
Am I wrong? Do I miss something?
answered Jul 31 at 18:33
md2perpe
5,7821922
I think you're right. I just happened to ask a question relating to the continuity of the extension of $f$ which of course requires that $overlinef$ is well-defined at $bin Bsetminus A$.
– TheLast Cipher
Aug 1 at 2:47
add a comment |Â
up vote
0
down vote
I don't see why it should be necessary to show that our constructed extension is well-defined, at least not because of different choices of sequences.
Define one extension $bar f : A to C$ by the following procedure:
For every $a in A$ set $bar f(a) = f(a).$ For every $b in B setminus A$ take one sequence $(a_k) subset A$ converging to $b$ (such a sequence exists by denseness of $A$) and set $bar f(b) = lim f(a_k).$ Here we need to show that the limit exists (this you might call well-definedness), but right now we don't need to show that it does not depend on the choice of sequence.
Now we have an extension. We have two steps left: 1) to show that $bar f$ is continuous, 2) to show that it is unique. One way to show the latter might be to show that it doesn't depend on the choices of sequences.
Am I wrong? Do I miss something?
answered Jul 31 at 18:33
md2perpe
5,7821922
I think you're right. I just happened to ask a question relating to the continuity of the extension of $f$ which of course requires that $overlinef$ is well-defined at $bin Bsetminus A$.
– TheLast Cipher
Aug 1 at 2:47
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I don't see why it should be necessary to show that our constructed extension is well-defined, at least not because of different choices of sequences.
Define one extension $bar f : A to C$ by the following procedure:
For every $a in A$ set $bar f(a) = f(a).$ For every $b in B setminus A$ take one sequence $(a_k) subset A$ converging to $b$ (such a sequence exists by denseness of $A$) and set $bar f(b) = lim f(a_k).$ Here we need to show that the limit exists (this you might call well-definedness), but right now we don't need to show that it does not depend on the choice of sequence.
Now we have an extension. We have two steps left: 1) to show that $bar f$ is continuous, 2) to show that it is unique. One way to show the latter might be to show that it doesn't depend on the choices of sequences.
Am I wrong? Do I miss something?
answered Jul 31 at 18:33
md2perpe
5,7821922
I don't see why it should be necessary to show that our constructed extension is well-defined, at least not because of different choices of sequences.
Define one extension $bar f : A to C$ by the following procedure:
For every $a in A$ set $bar f(a) = f(a).$ For every $b in B setminus A$ take one sequence $(a_k) subset A$ converging to $b$ (such a sequence exists by denseness of $A$) and set $bar f(b) = lim f(a_k).$ Here we need to show that the limit exists (this you might call well-definedness), but right now we don't need to show that it does not depend on the choice of sequence.
Now we have an extension. We have two steps left: 1) to show that $bar f$ is continuous, 2) to show that it is unique. One way to show the latter might be to show that it doesn't depend on the choices of sequences.
Am I wrong? Do I miss something?
answered Jul 31 at 18:33
md2perpe
5,7821922
answered Jul 31 at 18:33
md2perpe
5,7821922
answered Jul 31 at 18:33
md2perpe
5,7821922
answered Jul 31 at 18:33
md2perpe
5,7821922
5,7821922
I think you're right. I just happened to ask a question relating to the continuity of the extension of $f$ which of course requires that $overlinef$ is well-defined at $bin Bsetminus A$.
– TheLast Cipher
Aug 1 at 2:47
add a comment |Â
I think you're right. I just happened to ask a question relating to the continuity of the extension of $f$ which of course requires that $overlinef$ is well-defined at $bin Bsetminus A$.
– TheLast Cipher
Aug 1 at 2:47
I think you're right. I just happened to ask a question relating to the continuity of the extension of $f$ which of course requires that $overlinef$ is well-defined at $bin Bsetminus A$.
– TheLast Cipher
Aug 1 at 2:47
I think you're right. I just happened to ask a question relating to the continuity of the extension of $f$ which of course requires that $overlinef$ is well-defined at $bin Bsetminus A$.
– TheLast Cipher
Aug 1 at 2:47
add a comment |Â
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If you don't show that $f$ is well defined you won't be able to show that the function you have defined on $B$ is continuous. The answer to your second quesion is if you pick the 1st, 3rd, 5th ... elements in the sequence you get a desired subsequence. As far as your comment below is concerned, you will have to write down the proof of continuity to convince yourself that you have a problem.
– Kavi Rama Murthy
Jul 30 at 6:11
Why? Is $a_n$ not arbitrary?
– TheLast Cipher
Jul 30 at 6:12
I know picking elements from the sequence $f(x_i)$ would get me a subsequence. My question was why can we assume that its subsequence converges to $c$. To emphasize, my question was regarding its convergence to $c$. Thanks.
– TheLast Cipher
Jul 30 at 6:21
See the definition of $c$.
– Kavi Rama Murthy
Jul 30 at 6:27