# Proof Explanation: Showing an extension to a continuous function

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## Main Question or Discussion Point

I am reading Kaplansky's text on metric spaces and this part seems redundant to me. It was stated below (purple highlight) that we need to show that the convergence of $(f(a_n))$ to $c$ is independent of what sequence $(a_n)$ converges to $b$, when trying to prove the claim $f(b)=c$. The way I understood this part of the proof is, he has already shown that $f(b)=c$ when he chose an arbitrary sequence at the part where I underlined with red. So why is he repeating the argument after that?

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andrewkirk
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He has defined $f(b)$ to be the limit in $C$ of the sequence $\{f(a_i)\}$. So the limit depends on the choice of the sequence $\{a_i\}$. Hence it is conceivable that choice of a different sequence, say $\{\alpha_i\}$ would give a different limit, that is, that the limit of $\{f(\alpha_i)\}$ may be different from the limit of $\{f(a_i)\}$.

So he needs to prove that, provided $\{a_i\}$ and $\{\alpha_i\}$ both converge to $b$, the sequences $\{f(a_i)\}$ and $\{f(\alpha_i)\}$ both converge to the same point in $C$.

This process is called 'showing the function is well-defined'. Whenever we make a definition that uses an arbitrarily selected additional piece of information that is not in the notation for what we are defining, we need to show that the definition is well-defined, ie that it does not depend on the choice of that additional information.

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This process is called 'showing the function is well-defined'. Whenever we make a definition that uses an arbitrarily selected additional piece of information that is not in the notation for what we are defining, we need to show that the definition is well-defined, ie that it does not depend on the choice of that additional information.
Perfect response! I was not aware of this. Thank you!!

This process is called 'showing the function is well-defined'.
A follow up question. For me, it's not immediately evident that I need to show that the function $f$ is well-defined. How do I know when I need to show that a function $f$ is well-defined? I don't remember learning this in my introductory proof writing class.

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andrewkirk
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This may not cover all cases, but all the cases I can think of where one has to show that a function is well-defined is where a representative element of one of the items described in, or implied by, the function 'setup' plays a critical role in completing the definition. To be well-defined, it must be the case that the choice of the representative element doesn't give a different outcome.

In the above case, we are defining the extended function, which maps each point in B to a unique point in C. We consider a point b in B, and consider the set S of all sequences in A whose limit is b. We then choose from out of S a representative element $\{a_i\}$, and define $f(b)$ to be the limit in C of the sequence $\{f(a_i)\}$. But we can imagine different choices of representative elements giving us different points for $f(b)$. So we need to prove that doesn't happen - that if we choose any other sequence in S, applying the definition to that sequence will give us the same point.

Here is another example of definitions that need to be shown to be well-defined:

When working with groups in number theory, for any integer $m$ and a set of integers $G$ that is a group under the operation of addition, we define the 'm coset of G', denoted by $m+G$, as the set of integers $\{m+g\ :\ g\in G\}$.
So if $G=5\mathbb Z$ and $m=12$ the coset $m+G$, which is $12 +5\mathbb Z$ is the set $\{...,-8,-3,2,7,12,17,....\}$. Note that this is identical to $r+2\mathbb Z$ for any $r$ in the coset. For instance it's identical to $-8+5\mathbb Z$ and to $2+5\mathbb Z$. SO in our notation, the item before the $+G$ part is just a representative element of the coset.
We can then define an operation of addition between cosets of G by saying the sum of the cosets $a+G$ and $b+G$ is the coset $(a+b)+G$.
But a and b are just representative elements of the two cosets being added, so we need to prove that using different representative elements doesn't give a different coset for the sum. For instance we need to show that
$(-8+52\mathbb Z) + (4+5\mathbb Z)$, which is defined to be $((-8+4)+5\mathbb Z) = -4+5\mathbb Z)$, is equal to
$(2+5\mathbb Z) + (-1+5\mathbb Z)$, which is defined to be $((-8+4)+5\mathbb Z) = 1+5\mathbb Z)$.
It is not hard to prove that that is the case, but it needs to be done.

It is often not easy to spot when a proof of well-definition is needed. When confronted with a new definition, one needs to scrutinise the steps taken in producing the defined item, to see if there is any implicit or explicit choice made in there. In the case of the OP, it's fairly straightforward, as we explicitly choose a single sequence from the set of all sequences that converge to b. In the coset case it's less obvious, because the notation $-8+5\mathbb Z$ makes it look like $-8$ has been specified, but in fact it is just part of the notation and what it refers to is just the set $\{...,-8,-3,2,7,12,17,....\}$. The trick is to remember that sometimes choices are made in the notation that are not an intrinsic part of what is being defined. There is nothing special about -8 in the set $\{...,-8,-3,2,7,12,17,....\}$ and $-8+5\mathbb Z$ is just one of an infinite number of different possible ways of referring to it. $-8+5\mathbb Z$ is the same object as $2+5\mathbb Z$ so if we make any claim about that object, or a definition for some operation on it, we need to prove that the truth of the claim or the outcome of the definition does not depend on the choice of notation we make to refer to it.

We then choose from out of S a representative element $\{a_i\}$, and define $f(b)$ to be the limit in C of the sequence $\{f(a_i)\}$. But we can imagine different choices of representative elements giving us different points for $f(b)$. So we need to prove that doesn't happen - that if we choose any other sequence in S, applying the definition to that sequence will give us the same point.
Somehow, the way I understand when he said "there is a sequence $\{a_i\}$ that converges to $b$" is that $\{a_i\}$ is arbitrary in the set $S$ of all sequences in $A$ that converges to $b$. That's why I am under the impression that it's no longer needed to show that $f$ is well-defined. Apparently, I am wrong. So is $\{a_i\}$ arbitrary or not? Though it seems like you're implicitly saying that it isn't arbitrary and I don't understand why it isn't arbitrary. Thank you for the elaborate response.

andrewkirk
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@Terrell - it is arbitrary, but in proving the function is well-defined we show that it doesn't matter which arbitrary sequence we pick.

Here's an analogy. Say I want to measure the temperature of a big saucepan of boiling water. I pick a place in the water surface and plunge a thermometer in, then take it out, read the temperature and say 'that's the temperature of the water'. Now my choice of the place where to plunge the thermometer is arbitrary. So we should be concerned that the temperature reading will vary depending on which point I arbitrarily pick. Fortunately, using some information about the heat-conductivity of water and the rapidity of mixing in a pot of boiling water, I can prove that the temperature will hardly vary at all, no matter where I plunge the thermometer. So the choice of where to plunge it is arbitrary but it makes no material difference to the reading. The water temperature as defined by that method is 'well-defined'.

But we can imagine different choices of representative elements giving us different points for $f(b)$.
I think I glossed over this part a bit too quickly. When I was proving the proposition on my own, a voice in my head keeps telling me "what if $f(b)$ is a removable discontinuity and hence, undefined?".I think this could be "the why" I was looking for.

but in proving the function is well-defined we show that it doesn't matter which arbitrary sequence we pick.
I agree with this completely. I am just having issues on the necessity of showing $f$ is well-defined.

andrewkirk
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When I was proving the proposition on my own, a voice in my head keeps telling me "what if $f(b)$ is a removable discontinuity and hence, undefined?".I think this could be "the why" I was looking for.
Consider two cases, $b\in A$ and $b\in B-A$.

If $b\in A$ it can't be a discontinuity, removable or otherwise, because we are told that the original function $f:A\to C$ is uniformly continuous.

If $b\in B-A$ we need to prove that the extended function $f$ is continuous at $b$. I imagine he does that in a later part of the proof that is not shown above. He has to, because the Theorem states that the extended function is continuous on $B$. It will be a consequence of the way the extended function is constructed, and will use the fact that the original $f$ is uniformly continuous.

He has defined $f(b)$ to be the limit in $C$ of the sequence $\{f(a_i)\}$. So the limit depends on the choice of the sequence $\{a_i\}$.
Re-reading your first answer made it click for me. Thank you. You mentioned here that he defined $\lim \{f(a_i)\}=f(b)$. I tried to prove it on my own and I also didn't realize that I was defining it too. I now definitely believe this subtlety of defining something or not defining is what lead me to this ridiculous confusion. Thank you for your precise explanations!!

Math_QED