Measure zero (a possible problem in the reasoning)

[MathematicalPhysicist]

I read the proof here:
https://proofwiki.org/wiki/Measure_of_Empty_Set_is_Zero
that the measure of the empty set is zero.

 

[PeroK]

I read the proof here:
https://proofwiki.org/wiki/Measure_of_Empty_Set_is_Zero
that the measure of the empty set is zero.

My problem is with the last line of the proof where from the fact that: $\sum_{n=2}^\infty \mu(\emptyset) =0$ they conclude that $\mu(\emptyset) =0$.

Now here's my problem with this.

If we had some constant $c$ that we knew that it's identically zero then summing an infinite sum of zeros will end up in a zero (from the fact that the sequence of $\sum_{n=2}^k 0 \to 0$ as $k\to \infty$, but here we assume some constant that satisfies: $\sum_{n=2}^\infty c =0$, i.e. $\lim_{k\to \infty} \sum_{n=2}^k c = 0$, thus we get $c\infty =0$, so we conclude that $c=0/\infty$, which is not defined as a number.

We don't get this at all. If $c \ne 0$, we can constuct a contradiction with the infinite sum of $c$ being zero. Let $\epsilon > 0 \dots$.,

 

[sysprog]

I sympathize with your objections to the linguistic promiscuity of the use of the term 'zero' -- when I raised similar objections in calculus class, such as "why is it zero when we look at anyone slice and non-zero when we're summing them", the professor introduced me to the concept of 'treated as zero'.

 

[member 587159]

Usually $\mu(\emptyset) = 0$ is part of the definition of measure.

That being said, the theorem uses the statement that

$\sum_{n=0}^\infty c = 0 \implies c = 0$

for $c \in \mathbb{R}$.

This is rather straightforward to prove:

We have $\lim_{k \to \infty} (k+1)c = \lim_{k \to \infty} \sum_{n=0}^k c=0$

which is only possible if $c=0$ (otherwise this converges in absolute value to $+\infty$).

 

[jedishrfu]

One other point here is that infinity is not a number nor does it represent some infinitely large number.

Hence you can't use arithmetic operations on infinity and get anything meaningful out of them.

 

[MathematicalPhysicist]

Usually $\mu(\emptyset) = 0$ is part of the definition of measure.

That being said, the theorem uses the statement that

$\sum_{n=0}^\infty c = 0 \implies c = 0$

for $c \in \mathbb{R}$.

This is rather straightforward to prove:

We have $\lim_{k \to \infty} (k+1)c = \lim_{k \to \infty} \sum_{n=0}^k c=0$

which is only possible if $c=0$ (otherwise this converges in absolute value to $+\infty$).

If you use the arithematics of limits you get that: $\lim_{k\to \infty}(k+1)c = c\lim_{k\to \infty} (k+1)=c\infty$, so you get the equation $c\infty = 0$, which is not even defined.

 

[MathematicalPhysicist]

If for $c\infty = 0$ you get the solution $c=0$ then you actually define that $0/0 = \infty$, which even in limits in calculus is nonsense since we have L'hopital's rule for these kinds of limits.

 

[member 587159]

If you use the arithematics of limits you get that: $\lim_{k\to \infty}(k+1)c = c\lim_{k\to \infty} (k+1)=c\infty$, so you get the equation $c\infty = 0$, which is not even defined.

Your logic is flawed. That's not what I did. In the definition of limit, $+\infty$ is just a symbol. $\lim_{n\to \infty} a_n=+\infty$ just means that

$$\forall M\in\mathbb{R}: \exists n_0:\forall n\geq n_0: a_n\geq M$$

Basically, my proof used that it is impossible for a sequence to both converge to $0$ and $+\infty$.
_________

I don't want to confuse you,but in measure theory $\infty$ is often considered as a number. For example, in the definition of the Lebesgue integral, it is a convention that $0.\infty =0=\infty.0$! Care has to be taken at all times.

 

[PeroK]

If you use the arithematics of limits you get that: $\lim_{k\to \infty}(k+1)c = c\lim_{k\to \infty} (k+1)=c\infty$, so you get the equation $c\infty = 0$, which is not even defined.

There are several things here.

1) This is wrong, as @Math_QED has pointed out. Limits are defined by so-called "epsilon-delta" definitions, as part of a branch of mathematics called Real Analysis.

2) If you haven't learned Real Analysis you are not going to be able to learn or understand Measure Theory.

3) In Real Analysis, $\infty$ is not a number, but it is used as a shorthand as follows:

We write $\lim_{n \rightarrow \infty} a_n = +\infty$ as a shorthand for the sequence $a_n$ diverges to $+\infty$ as $n \rightarrow \infty$. In either case this means that:
$$\forall \ K > 0, \ \exists n_0, \ \text{such that} \ n > n_0 \ \Rightarrow \ a_n > K$$

4) Nothing in Real analysis demands working with $\infty$ as a number or having the properties of a number. However, you will see things like:

Let $\lim a_n = L$, where $L$ is a real number or $\pm \infty$. That again is a shorthand for three separate cases: the limit is some finite number $L$, the sequence diverges to $+\infty$, and, the sequence diverges to $-\infty$.

5) Anything short of this rigorous attention to detail won't do if you want to learn Measure Theory.

 

[sysprog]

Great post, @PeroK

 

[MathematicalPhysicist]

Your logic is flawed. That's not what I did. In the definition of limit, $+\infty$ is just a symbol. $\lim_{n\to \infty} a_n=+\infty$ just means that

$$\forall M\in\mathbb{R}: \exists n_0:\forall n\geq n_0: a_n\geq M$$

Basically, my proof used that it is impossible for a sequence to both converge to $0$ and $+\infty$.
_________

I don't want to confuse you,but in measure theory $\infty$ is often considered as a number. For example, in the definition of the Lebesgue integral, it is a convention that $0.\infty =0=\infty.0$! Care has to be taken at all times.

I don't understand why you both think that I didn't learn calculus 1-3 or for that matter don't have a BSc degree in maths and physics?!

I did learn measure theory a few years ago (in my country most students don't really learn, they just learn to pass the exams so most of my years in university were spent from one strike to another so I had the time to learn most my education from books by myself, also for the courses so in the end I learned a little bit of measure theory from some book from the open university), and I didn't object then to the definition of the new number $\infty$ and the two equalities you stated.

But now when I saw this proof from the link I've posted in my OP I have some reservations.

I mean is the infinity symbol defined in measure theory as you stated the same as to the notion from classical calculus (i.e Cauchy's, Weistrauss's formalism)?

Because it seems the sum in the proof over there does use the notion of limit from calculus, is it not?
Now that I am older than back then when I read the book I think it's dubious to define it like that.

 

[sysprog]

I don't understand why you both think that I didn't learn calculus 1-3 or for that matter don't have a BSc degree in maths and physics?!

I did learn measure theory a few years ago (in my country most students don't really learn, they just learn to pass the exams so most of my years in university were spent from one strike to another so I had the time to learn most my education from books by myself, also for the courses so in the end I learned a little bit of measure theory from some book from the open university), and I didn't object then to the definition of the new number $\infty$ and the two equalities you stated.

But now when I saw this proof from the link I've posted in my OP I have some reservations.

I mean is the infinity symbol defined in measure theory as you stated the same as to the notion from classical calculus (i.e Cauchy's, Weistrauss's formalism)?

Because it seems the sum in the proof over there does use the notion of limit from calculus, is it not?
Now that I am older than back then when I read the book I think it's dubious to define it like that.

I think that no-one here called into question your credibility -- maybe someone who is more strongly than you are a proponent of someone's version of measure theory didn't take or at least didn't express the same view as you did -- when things get to 'calculus and beyond', then some competent, some brilliant, and some outlying mathematicians (or some of those who are any of the possible combinations thereof) might not always agree with one another.

 

[sysprog]

Oh and, for my part, I vote for Cauchy.

 

[Stephen Tashi]

If you use the arithematics of limits you get that: $\lim_{k\to \infty}(k+1)c = c\lim_{k\to \infty} (k+1)=c\infty$, so you get the equation $c\infty = 0$, which is not even defined.

Proceeding correctly, you don't get that equation unless you use theorem that says you can apply the arithmetic of limits to such situations. If you use such a theorem, it doesn't make the general claim that "$0 \cdot \infty = 0$" . ( A theorem about the specific case $lim_{k \rightarrow \infty} (kc)$ is not proven by assuming that such arithmetic. is correct or even has a defined meaning. It must be proven using definitions of limits. )

The "arithmetic of limits" has limited applicability. It is an abbreviation for statements about limits. The notation used in the arithmetic may suggest manipulations with numbers, but properly interpreted, it does not assume $\infty$ represents a number. Likewise the statement $lim_{x \rightarrow a} f(x) = \infty$ is, technically, not a statement about the limit of a function, but rather a statement of a particular way in which the limit of a function can fail to exist. Mathematics must admit to using confusing terminology. From a literary point of view, the terminology "infinite limit" is self-contradictory. The standard defintion of $lim_{x \rightarrow a} f(x) = L$ applies only the situation where $a$ and$L$ are real numbers. Describing different situations that we call an "infinite limits" cannot be done by changing the words of the standard definition and letting $L$ or $a$ "be infinity". Instead, it requires a complete re-write.

(The literary approach to interpreting the effect of adjectives on mathematical definitions doesn't work. For example, in literature, "a big X" can safely be assumed to be "an X that is big". In mathematics "a big X" might be defined as something that is not an X at all. )

If you are reading a respectable text, the author will expect the reader to supply a proper proof that $lim_{k \rightarrow \infty} (kc) = 0$ implies $c = 0$. A proper proof won't be based on the incorrect general idea that if $lim_{k \rightarrow \infty} f(k) = \infty$ and $lim_{k \rightarrow \infty} g(k) = 0$ then $lim_{k \rightarrow \infty} (f(k) g(k)) = \infty \cdot 0 = 0$.

 

[WWGD]

I sympathize with your objections to the linguistic promiscuity of the use of the term 'zero' -- when I raised similar objections in calculus class, such as "why is it zero when we look at anyone slice and non-zero when we're summing them", the professor introduced me to the concept of 'treated as zero'.

I know it's frustrating and you have a valid point but unless you use some shorthand and cut a few corners, dealing with all these issues will likely go on for ever in a chaotic way.

 

[sysprog]

@Stephen Tashi in re your post #14 in this thread great post Sir . . .

 

[WWGD]

@Stephen Tashi in re your post #14 in this thread great post Sir . . .

See? Don't mean to beat my point to death but look at all Stephen had to write to clarify a single point, to unpack the shorthand used. Now multiply that by the number of similar issues. Too long to explain . Basically, details left to the reader and come up with any questions.

 

[sysprog]

@WWGD I think that this remark in post #14 by @Stephen Tashi is beautifully ironic:

The "arithmetic of limits" has limited applicability.

 

[MathematicalPhysicist]

@Stephen Tashi my problem is with adding the number infinity and defining its multiplication with zero before defining the limits to infinity. We might as well define this number to satisfy: $\infty \cdot 0 =0\cdot \infty =\infty$ and build a different theory from Lebesgue's and Borel's theory.

 

[member 587159]

@Stephen Tashi my problem is with adding the number infinity and defining its multiplication with zero before defining the limits to infinity. We might as well define this number to satisfy: $\infty \cdot 0 =0\cdot \infty =\infty$ and build a different theory from Lebesgue's and Borel's theory.

What do you hope to gain by making such a definition? The measure theoretic convention is that $0.\infty=0=\infty.0$.

Otherwise you get problems with defining product measures (and more things).

I politely suggest to take a step back and master real analysis,then come back to measure theory.

 

[Stephen Tashi]

@Stephen Tashi my problem is with adding the number infinity and defining its multiplication with zero before defining the limits to infinity.

(Standard) mathematics does not take this approach. Are you thinking otherwise?

Hearing "Everything you ever learned is wrong" is a familiar experience for students! Transitioning from other topics in the liberal arts to mathematics is a good example. In the liberal arts we are advised "To understand the meaning of a sentence, break it up into phrases and consider the meaning of the individual words in the phrases". In mathematics, the individual words in a sentence may have no defined meaning, so you can't reliably deduce the meaning of a sentence in that manner. Likewise, in mathematical notation, the individual symbols may have no defined meaning, only the total aggregation of symbols into a particular pattern has an interpretation.

The definition of the notation $\sum_{k=1}^{\infty} f(k)$ does not assume or define a number denoted by "$\infty$". The definition of the notation "$\sum_{k=1}^{\infty} 0$" does not imply that a number called "infinity" is multiplied by zero , nor does it imply that zero is added "an infinite number of times". (Nor is any definition of "infinity" as a number incorporated in the definitions for those notations.)

People can form their own private notions about what formal mathematical definitions mean, but these private notions are not the content of mathematics. People can use poetic language (almost everyone does sometimes) and describe $\sum_{k=1}^\infty 0$ by the lurid phrase "adding zero infinitely many times", but such descriptions are not mathematical definitions.

For example, the notation "$\sum_{k=1}^{\infty} f(k)=L$" is an abbreviation for the sentence "The limit of the function $g(k) = \sum_{i=1}^k f(i)$ as k approaches infinity is equal to L". However, the mathematical definition of this sentence does not define the individual words "limit", "approaches" or "infinity". Nor does it assume the meaning of that sentence is created by having definitions of those individual words.

Mathematical definitions define complete sentences ( that use terms that may be undefined) as being equivalent to other complete sentences (who terms are defined). The complete sentence: "The limit of the function g(k) as k approaches infinity is equal to L" is defined by saying it is equivalent to the complete sentence "For each number epsilon greater than zero there exists a integer N such that if k is a number greater than N then the absolute value of the difference between the number L and g(k) is less than epsilon". In this definition, no meaning is defined or assumed for the word "infinity".

 

[MathematicalPhysicist]

What do you hope to gain by making such a definition? The measure theoretic convention is that $0.\infty=0=\infty.0$.

Otherwise you get problems with defining product measures (and more things).

I politely suggest to take a step back and master real analysis,then come back to measure theory.

I pointed out that in the proof of my OP why I think there's a flaw in the proof here:
https://proofwiki.org/wiki/Measure_of_Empty_Set_is_Zero

where they take a sum of infinite number of terms, either they use the notion of the symbol that satisfies $\infty \cdot 0 = 0$ or they use the notion of limits of sums, in which case we have the following equation:
$\sum_{n=1}^\infty c = \lim_{k\to \infty}\sum_{n=1}^k c = c\lim_{k\to \infty}k=c\infty = 0$, as far as I see they use both notions of this symbol.
Let me know where do you object to my reasoning here?

As far as I can tell they use both notions of this symbol $\infty$, part as the limit to infinity and part as the external number beyond the real numbers that satisfy the extra condition $\infty = 0\cdot \infty$ which in the limits interpretation which is inconceivable.

I assume if they make this error elsewhere in measure theory, that you bound to get an inconsistent theory.

 

[MathematicalPhysicist]

What I am trying to say is that if we know that $c=0$ then an infinite sum of zero will give you zero.
But if you are given a sum of an infinite number of $c$'s and equate it to zero you cannot know that $c=0$ because this sum doesn't converge.

 

[member 587159]

Okay, I think continuing this discussion is fruitless. I'm going to provide you an explanation that does not use the extended number $\infty$ at any point.

Claim: Let $a$ be a real number with $\sum_{k=1}^\infty a = 0$. Then $a=0$.
Proof: The assumption implies that the sequence $(na=\sum_{k=1}^n a)_{n=1}^\infty$ converges. But the only value for which this happens is $a=0$, as $(na)_{n=1}^\infty$ is an unbounded sequence for $a \neq 0. \quad \square$

Does this solve your problem?

 

[Stephen Tashi]

I pointed out that in the proof of my OP why I think there's a flaw in the proof here:
https://proofwiki.org/wiki/Measure_of_Empty_Set_is_Zero

As far as I can tell they use both notions of this symbol $\infty$,

None of the manipulations you gave using the symbol $\infty$ appear in that link. So it isn't "they" who offer the arguments that you have invented.

I think what you want is a proper justification from going from the statement: $\sum_{k=1}^\infty \mu(\emptyset) = 0$ to the conclusion $\mu(\emptyset) = 0$. It's a fair criticism to say that the link does not prove this conclusion. Your criticisms are directed toward your own attempts to justify it.

@Math_QED has indicated a proof if you accept the last line in it. If we want more detail, I think it will involve a proof by contradiction and it will require knowing the formal definition of $\lim_{k \rightarrow \infty} g(k) \ne L$

 

[MathematicalPhysicist]

The formal definition of $\lim_{k\to \infty}g(k)\ne L$ is simply $\exists \epsilon >0 \forall N \exists k>N: |g(k)-L|\ge \epsilon$; how would you proceed from here.

OK, I see my error, thanks for pointing it out to me all of you.

Sorry to be stubborn, just when you get older you get to be stubborner.

But I agree if $c \ne 0$ then we get $\pm \infty = 0$ which is obviously a contradiction, thus $c=0$.
So we get: $\lim c(k+1)= \lim 0(k+1)=0$.

So foolish of me... :-(
Is there an option to delete this thread? I don't see an option somewhere.

 

[Stephen Tashi]

The formal definition of $\lim_{k\to \infty}g(k)\ne L$ is simply $\exists \epsilon >0 \forall N \exists k>N: |g(k)-L|\ge \epsilon$; how would you proceed from here.

An outline of a proof-by-contradiction;
Assume $\sum_{k=2}^\infty \mu(\emptyset) = 0$ and $\mu(\emptyset) \ne 0$

Case 1: $\mu(\emptyset) = c > 0$
Show $\sum_{k=2}^\infty c \ne 0$
Let $\epsilon = 1$. For a given $N$ let $k$ be the larger of $N+1$ and the greatest integer in $100/c$. Show $|\sum_{i=2}^k c - 0| > \epsilon$.

Case 2: $\mu(\emptyset) = c < 0$
Show $\sum_{k=2}^\infty \ne 0$.
Let $\epsilon = 1$ For a given $N$ let $k$ be the larger of $N+1$ and the greatest integer in $100/|c|$. Show $|\sum_{i=2}^k c - 0| > \epsilon$.

 

[MathematicalPhysicist]

I have a further question about the definition of the symbol $\infty$ in measure theory, if we define it to satisfy: $\infty \cdot 0 = 0\cdot \infty = 0$ doesn't it mean that: $0/0=\infty$ for this symbol?

In which key results in measure theory do they use the multiplication condition?

 

[member 587159]

I have a further question about the definition of the symbol $\infty$ in measure theory, if we define it to satisfy: $\infty \cdot 0 = 0\cdot \infty = 0$ doesn't it mean that: $0/0=\infty$ for this symbol?

In which key results in measure theory do they use the multiplication condition?

Compare with $1.0=0=0.1\implies 1=0/0$? Even in the real case you wouldn't make such a claim.

 

[Stephen Tashi]

I have a further question about the definition of the symbol $\infty$ in measure theory,

The symbol $\infty$ is not defined in the standard approach to measure theory. It appears as an element in patterns of symbols such as "$\cup_{i=1}^\infty s_i$", or "$lim_{x \rightarrow \infty} g(x) = \infty$". Such patterns of symbols are defined, but "$\infty$" is not defined in isolation.

Informal jingles such as "$\infty \cdot \infty = \infty$" are used to remember theorems such as: If $lim_{x\rightarrow a} f(x) = \infty$ and $lim_{x \rightarrow a} g(x) = \infty$ then $lim_{x \rightarrow a} (f(x) g(x)) = \infty$. Such theorems do not assert that $\infty$ by itself is a object that obeys certain arithmetic.

if we define it to satisfy: $\infty \cdot 0 = 0\cdot \infty = 0$

There is no such definition in measure theory. The formal statement about limits implied by the jingle "$0 \cdot \infty = 0$" is not, in general, true.

 

[MathematicalPhysicist]

Compare with $1.0=0=0.1\implies 1=0/0$? Even in the real case you wouldn't make such a claim.

But $\infty$ isn't a real number.

In the real number case you would divide by $1$ for example. So you say that this symbol satisfies also: $0 = 0/\infty$?

 

[MathematicalPhysicist]

The symbol $\infty$ is not defined in the standard approach to measure theory. It appears as an element in patterns of symbols such as "$\cup_{i=1}^\infty s_i$", or "$lim_{x \rightarrow \infty} g(x) = \infty$". Such patterns of symbols are defined, but "$\infty$" is not defined in isolation.

Informal jingles such as "$\infty \cdot \infty = \infty$" are used to remember theorems such as: If $lim_{x\rightarrow a} f(x) = \infty$ and $lim_{x \rightarrow a} g(x) = \infty$ then $lim_{x \rightarrow a} (f(x) g(x)) = \infty$. Such theorems do not assert that $\infty$ by itself is a object that obeys certain arithmetic.
There is no such definition in measure theory. The formal statement about limits implied by the jingle "$0 \cdot \infty = 0$" is not, in general, true.

Well, @Math_QED said in post number 8 (I see the irony :-)) in this thread that in Measure theory that it's indeed satisfy this relation.

BTW, for those who don't see the irony I remind you of the following joke:

 

[member 587159]

There is no such definition in measure theory.

I beg to disagree. Here are standard references for measure theory that all mention it.

(1) Donald Cohn's "Measure theory" -> Appendix A, p380
(2) Follands "Real analysis" -> p11
(3) Rudin's "Real and complex analysis" -> p18

Indeed, such a definition is useful in situations like this:

Consider the product $\mu \otimes \nu$ of two ($\sigma$-finite measures) $\mu, \nu$. Such measures are completely determined by the property $\mu\otimes \nu(A \times B) = \mu(A) \nu(B)$.

Now, consider the case where $A = \emptyset, \nu(B) = +\infty$. Then the above condition reads $\mu\otimes \nu(\emptyset) = \mu\otimes \nu(\emptyset \times B) = \mu \otimes \nu (A \times B) = \mu(\emptyset) \nu(B) = 0 . \infty$ and the definition $0. \infty = 0$ is needed naturally, unless you always want to specify the exceptions.

You literally never run into trouble, as long as you don't make up rules yourself.

Really confused as why someone would say that things like $\infty + \infty$ are not defined in measure theory. Not only are they rigorously defined, they are also necessary!

For example, consider the counting measure $\sharp$ on $\mathbb{R}$. Then

$\sharp(\mathbb{Q} \cup \mathbb{Q}^c) = \sharp(\mathbb{Q})+ \sharp(\mathbb{Q}^c)= \infty + \infty$

Oh no! It's $\infty + \infty$! We are doomed! Unless we, maybe, define $\infty + \infty := \infty$? It is pure nonsense to say that extended reals are not rigorously defined. One just has to be careful with it. Context is everything.

 

[WWGD]

The symbol $\infty$ is not defined in the standard approach to measure theory. It appears as an element in patterns of symbols such as "$\cup_{i=1}^\infty s_i$", or "$lim_{x \rightarrow \infty} g(x) = \infty$". Such patterns of symbols are defined, but "$\infty$" is not defined in isolation.

Informal jingles such as "$\infty \cdot \infty = \infty$" are used to remember theorems such as: If $lim_{x\rightarrow a} f(x) = \infty$ and $lim_{x \rightarrow a} g(x) = \infty$ then $lim_{x \rightarrow a} (f(x) g(x)) = \infty$. Such theorems do not assert that $\infty$ by itself is a object that obeys certain arithmetic.
There is no such definition in measure theory. The formal statement about limits implied by the jingle "$0 \cdot \infty = 0$" is not, in general, true.

I remember Royden working under that assumption.

 

[Stephen Tashi]

I beg to disagree. Here are standard references for measure theory that all mention it.

We have to distinguish between defining something versus mentioning it . -

Using symbol as part of an aggregation of other symbols is mentioning it. This doesn't imply that a definition for the entire aggregation also gives a definition for the individul symbols that compose it.

I agree that the symbol "$\infty$" appears in writings on measure theory. I do not agree that this symbol has one specific definition that applies in all the places it appears.

Likewise, symbols such as "$\rightarrow$" appear in writings on measure theory. In that sense they are mentioned, but not given a definition in their own right.

(3) Rudin's "Real and complex analysis" -> p18

which is a book, I happen to have. Rudin says on p18

For any $E \subset X$, where $X$ is any set define $\mu(E) = \infty$ if $E$ is an infinite set and let $\mu(E)$ be the number of of points in $E$ if $E$ is finite.

This is not a definition of "$\infty$".

It could be considered to define the collection of symbols "$\mu(E) = \infty$" to mean "$E$ is an infinite set".

Prior to that, on page 8, Rudin uses the symbol "$\infty$" in the aggregation of symbols "$f:X \rightarrow [-\infty, \infty]$" , so if he's writing mathematics properly, he wouldn't used a concept on page 8 and then wait till page 18 to define it.

I think Rudin takes for granted that the reader will interpret $\infty$ in the context of the "extended real number line" in some places in his text. However, the arithmetic of an abstract number line does not explain why aggregations of symbols such as "$lim_{x \rightarrow a} f(x) = L$" and "$lim_{x \rightarrow a} f(x) = \infty$" require different definitions, one with the condition $|f(x) - L| < \epsilon$ and the other with the condition $f(x) > r$.

Indeed, such a definition is useful in situations like this:

However, I see no definition of "$\infty$" that is the same definition in all situations.

One just has to be careful with it. Context is everything.

I agree that the meaning of the symbol "$\infty$" is dependent on the context in which it used. My point is that it has no universal definition.