1
$\begingroup$

I was just reading from my calculus textbook. While reading the continuity part containing Intermediate Value Theorem, it says that the proof of IVT requires proof of completeness property of Real numbers first. I was just thinking that while studying the limits of functions, it is required that 'function should be defined around the limit point'. Is not this definition require the completeness property of Real numbers already?

So my question is, "Is not it necessary to prove the completeness of Real numbers before defining the limit of a function?"

Edit: (I was thinking like that):

To ensure that a function is continuous at a point, following conditions are required:

  1. Limit exists at that point.

  2. function is defined at that point.

  3. function value is equal to the limit.

If at a point inside the Domain of a function a limit exists and function value is equal to the limit at that point, then it ensures that function value at that point will not jump suddenly as compared to the function values at the adjoining points.

Now take $\mathbb{Q}$ as the domain of a function. Taking standard example in the textbook:

$$f(x) = 3x^2-1$$

Book says that IVT is not valid for this function (even after the function is continuous in $\mathbb{Q}$) because $\mathbb{Q}$ is incomplete and contains holes. For example, at point f(x) = 0, even if "0" is in $\mathbb{Q}$, but it is not in the range of the function.

But i think that stating completeness (by saying that function is continuous) is not required for this failure. Rather, it can be said that function f(x) is not continuous in $\mathbb{Q}$. Because limit does not exist for the points in $\mathbb{Q}$ adjoining $\sqrt{1/3}$ (as the function value suddenly jumps suddenly leaving the point f(x) = 0). So, the function fails to be continuous in $\mathbb{Q}$ (for the above given definition of continuity).

Therefore, i was thinking that if a function is continuous inside a number system, then it should follow that IVT will be valid for that function, even if the number system is complete or incomplete.

Also the jump discontinuity can have one of these two reasons:

a. Function is discontinuous at the point.

b. The number system in which the function is defined is incomplete (contains holes).

"Now is that reasoning correct or i am stating something incorrectly?"

$\endgroup$
11
  • 1
    $\begingroup$ You can define limits for functions with domains that are not complete; you do not need completeness for the definition of limits. What the definition of limit you refer to requires is just that every number “sufficiently close to $a$” be in the domain, but it doesn’t actually tell you that there are no “gaps” between the numbers. The definition would work just as well if the only numbers that “existed”, as far as the definition is concerned, were rationals. There are some results that may no longer work (e.g., increasing+bounded implies a limit exists), though. $\endgroup$ – Arturo Magidin Apr 4 '20 at 4:58
  • $\begingroup$ @ArturoMagidin Thanks for the comment, it is difficult to visualize but i can accept what you said. Now can you answer me why do statement of IVT requires completeness of Real numbers? $\endgroup$ – Harwinder Singh Apr 4 '20 at 6:07
  • 1
    $\begingroup$ You should have a look at this answer. Also consider yourself lucky as few intro calculus textbooks mention completeness property of reals. $\endgroup$ – Paramanand Singh Apr 4 '20 at 15:21
  • 1
    $\begingroup$ Also check another related answer. In short there is no calculus without completeness of reals. $\endgroup$ – Paramanand Singh Apr 4 '20 at 15:24
  • 1
    $\begingroup$ Btw the statement of IVT does not require completeness, but its proof requires it. Try to prove IVT and you will see things are not as obvious as they appear. In fact completeness is the secret, mysterious ingredient of calculus which most textbooks leave out. If you manage to grasp completeness then most of calculus and some intro real analysis is child's play. $\endgroup$ – Paramanand Singh Apr 4 '20 at 15:28
1
$\begingroup$

Imagine that you don’t know about the irrational numbers. All you know are rational numbers. You can still define rational-valued functions of rational variable (functions whose domain is a subset of $\mathbb{Q}$, and whose range is a subset of $\mathbb{Q}$). Let me call these “r-function” (unfortunately, “rational function” has a different meaning).

When we talk about “open intervals” in this setting, we just means the rational numbers there; so the open interval $(a,b)$ with $a,b$ rationals means $\{q\in\mathbb{Q}\mid a\lt q\lt b\}$.

You can also still define limits for such r-functions, exactly the same way you are used to, in this setting.

Definition. Let $f$ be an r-function, let $a\in\mathbb{Q}$. Assume that there exists an open interval containing $a$ such that $f$ is defined at every rational on that open interval. We say that the limit of $f$ as $x$ approaches $a$ is $L$ if and only if for every $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that if $x\in\mathbb{Q}$ and $0\lt|x-a|\lt \delta$, then $|f(x)-L|\lt \epsilon$.

The limit theorems will still be valid in this setting.

We can also define continuity in this setting:

Definition. Let $f$ be an r-function and let $a$ be a rational. We say that $f$ is *continuous at $a$ if and only if three things happen:

  1. $f$ is defined at $a$.
  2. The limit of $f$ as $x$ approaches $a$ exists.
  3. The value of the limit is $f(a)$.

Again, a lot of the theorems you know about limits and about continuity are still valid in this setting. Also, a lot of the functions you are familiar with that make sense in this setting will be continuous everywhere: polynomials, viewed as r-functions, are continuous everywhere (that is, at every $a\in\mathbb{Q}$).

It would be a good exercise for you to check what theorems go through.

Now here is the important part: you don’t know what the reals are. You do not imagine these functions as “living” on the real line and full of holes; because we are imagining a world where the non-rational real numbers simply do not exist, period. So when you imagine these functions, something like $f(x) = 3x^2-1$ is continuous everywhere. There is no such thing as $\sqrt{1/3}$. At every rational $a$, you can prove that $f(x)$ is continuous at $a$, under the definition given above.

But in this setting, the Intermediate Value Theorem is not valid. Because, for example, even though $f$ is continuous on $[0,1]$, and $f(0)\lt 0$ and $f(1)\gt 0$, there is no rational $q$ (which, remember, are the only numbers that exist here) in $(0,1)$ satisfies $f(q)=0$.

Why does this happen? Because the rationals are not complete. They fail to satisfy a key property that the reals do satisfy. It’s not that they have “holes” when you think of them as sitting inside the reals, but that they intrinsically fail to satisfy the completeness property. Namely, the following statement:

If $S$ is a nonempty subset of XXXX that is bounded above, then $S$ has a least upper bound in XXXX.

When XXXX is replaced by “the real numbers”, the statement is true. That is the Completeness of the Real Numbers. When you replace XXXX with “the rational numbers”, the statement is false. This is why we say that rational are not complete.

The fact that this fails is why the IVT fails for r-functions, even though one can certainly define limits and continuity for r-functions.

In fact, the IVT is equivalent to completeness: if you assume the real numbers are complete, you can prove the IVT. And if you do not assume the real numbers are complete, but you assume the IVT holds, then you can prove that the reals are complete from that hypotheses.


The definitions above can be used for real functions, and it doesn’t matter whether the reals are complete or not. You can see that because by changing “real” to “rational” everywhere, you still get a sensible definition that lets you prove the results about limits, even though the rationals are not complete. So you do not need the fact that the reals are complete to define limits, or open intervals, or neighborhoods, or continuity. All of those can be defined, and statements about them can be proven, whether or not the reals (or whatever numbers you choose to work with) are complete. That’s why we do not say that completeness is required for those definitions.

In calculus, the first time that completeness really comes in is when you prove the IVT. Up until that point, you never really need the completeness (though sometimes it is informally invoked to give intuition about definitions, such as the definition of continuity being an attempt at capturing the idea of “no holes, no jumps, no breaks”). But, formally, you do not need completeness for any of those definitions, or any of those theorems.


You say, with respect to $f(x)=3x^2-1$:

Rather, it can be said that function $f(x)$ is not continuous in $\mathbb{Q}$. Because limit does not exist for the points in $\mathbb{Q}$ adjoining $\sqrt{1/3}$ (as the function value suddenly jumps suddenly leaving the point $f(x) = 0$). So, the function fails to be continuous in $\mathbb{Q}$ (for the above given definition of continuity).

That’s wrong.

First, what does “points adjoning $\sqrt{1/3}$” mean? Give my any rational $q$, and the function satisfies the definition of limit at that point, with limit equal to $q^2-1=f(q)$. The function is continuous at any $q$, no matter what $q$, no matter how close it is to the real number $\sqrt{1/3}$. As an r-function, there are no “sudden jumps” in the sense you seem to think there are.

Remember that continuity is not about pictures, not about “holes”. It’s about the definition. And the definition is satisfied, so the function is continuous. The “number” $\sqrt{1/3}$ may as well not exist as far as the r-function is concerned.

Second “the point $f(x)=0$”. That’s not a point in the domain. The fact that the function does not take the value is the failure of the Intermediate Value Theorem, not of continuity. Because continuity is connected to the IVT only because you have completeness. Without completeness, the two are not connected.

That’s why we have a definition of continuity, one that is mathematical, and not merely an intuitive notion of “no jumps”.

(Recall that the definition of continuity still yields results and that are sometimes hard to swallow, like the fact that the function $f(x) = 0$ if $x\in\mathbb{Q}$ and $f(x)=x$ if $x$ is irrational is in fact continuous at $0$, even though it is full of holes and jumps...)


What is going on is that there is a much more general definition of continuity and of limits than the one you are seeing; but one that would be too general and abstract for most students at your level. A definition that relies on the notion of “closeness”, “neighborhoods”, “open sets”. They are studied as part of a field called “topology”, and the case of the real numbers is just one particular instance of these general notions, one where you have a bunch of other tools that make things easier and more concrete.

There is no surprise that you are getting confused, given that you are trying to push ideas that have been simplified and anchored in the real numbers to places where they no longer make full sense without being able to abstract and completely divorce yourself from those anchors. Which kind of defeats the purpose of putting those anchors down in the first place (which are there to help ground things as you learn them for the first time). These are not easy things to do on a first try.

$\endgroup$
1
  • $\begingroup$ Thanks for the comment. I can understand some of the mistakes pointed out by you. It was a misconception that limit does not exist for some points on this function. Because we can prove the continuity of this function based upon the definition of continuity. But i think, i should read further to completely understand what you said and to get rid of those simplified notions you pointed out. Thanks again. $\endgroup$ – Harwinder Singh Apr 6 '20 at 5:10
1
$\begingroup$

Consider your example $f:\mathbb {Q} \to\mathbb {Q} $ defined by $$f(x) =3x^2-1$$ It is defined everywhere in $\mathbb {Q} $. And one can prove (using a suitable definition of continuity) that $f$ is continuous on $\mathbb {Q} $.

The definition of continuity which you have provided works well here. Consider an arbitrary point $a\in\mathbb {Q} $. Then $f(a) =3a^2-1$. We can show that $$\lim_{x\to a} f(x) =3a^2-1=f(a)$$ easily. Note that $$|f(x) - f(a) |=3|x+a||x-a|$$ and let $\epsilon \in\mathbb {Q}, \epsilon>0$ be given. If $|x-a|<1$ then $$|x+a|\leq |x-a|+2|a|<1+2|a|$$ and hence $$|f(x) - f(a) |<3(1+2|a|)|x-a|$$ and thus if $\delta=\min(1,\epsilon/(3+6|a|))$ then $$0<|x-a|<\delta\implies |f(x) - f(a) |<\epsilon $$ and therefore $f$ is continuous at $a$.

Clearly IVT does not work for this function as we have $f(0)=-1,f(1)=2$ and yet there is no $a\in\mathbb {Q} $ for which $f(a) =0$.

Your misconception stems from the false belief that $f(x)$ jumps at points near $\sqrt{1/3}$ and leaves out $0$. Well there are no such jumps in values of $f$ (because it satisfies the definition of continuity as shown above). Consider: if there is jump what is size of this jump? The function $f(x) $ never vanishes but it takes values as close to $0$ as we please. Thus given any positive rational $b$ there are rationals $a, a'$ such that $$-b<f(a') <0<f(a)<b$$ Try to prove this for yourself using properties of rational number system. Don't use things like $\sqrt{1/3}$.


Your belief is essentially based on a failure to distinguish between two properties density and completeness. Consider the system of integers. It has obvious gaps. There is no integer between $1$ and $2$ and the gap between $1$ and $2$ is of size $1$.

The system of rationals has no obvious gaps. Given two rationals there is always another one lying between them no matter now close the two given rationals are. This property goes by the name density. Integers are not dense but rationals are dense. The range of your function $f$ is also dense (prove this).

It is only because of density of rationals that it is difficult to visualize their gaps. Here the size of gap is $0$. Well it apppers something is fishy here. If the gap size is $0$ then it means there are no gaps.

The right way to visualize gaps in rationals was provided by Dedekind. He said that if the rationals are partitioned into two sets $A, B$ such that every number in $A$ is less than every number in $B$ then there should be a rational which acts as a boundary between $A$ and $B$. If $A$ consists of all rationals less than or equal to $1$ and $B$ consists of rationals greater than $1$ then $1$ acts as a boundary point between $A$ and $B$.

Dedekind constructed a partition of $\mathbb {Q} $ into sets $A, B$ such that every member of $A$ is less than every member of $B$ and yet there is no boundary between $A$ and $B$ and this is what Dedekind presented as conclusive evidence of gaps in rational number system. The following sets $$A=\{x\mid x\in\mathbb {Q}, x>0,x^2<2\}\cup \{x\mid x\in\mathbb {Q}, x\leq 0\},B=\mathbb{Q} \setminus A$$ are the most popular example of such a partition.

On the other hand the system of real numbers possesses completeness in the sense that any partition of the real numbers into two sets as explained above leads to a boundary point.

$\endgroup$
3
  • $\begingroup$ Thanks for the detailed answer (edited portion is also nice). As far as i can understand, what you are trying to say is that rational numbers can satisfy continuity property due to their density properties but are unable to satisfy IVT and other important theorems due to their incompleteness. This makes sense to me, but i will not pretend to completely understand what you said. I think, first i should read your referred text (G.H. Hardy) and then try to reason over the answers given by both of you. Thanks again. $\endgroup$ – Harwinder Singh Apr 6 '20 at 5:02
  • $\begingroup$ @HarwinderSingh: Hardy's book is available online free of cost in djvu format if you search enough. Also go for 10th edition and not previous editions. $\endgroup$ – Paramanand Singh Apr 6 '20 at 8:14
  • $\begingroup$ @HarwinderSingh: yes density of rationals is key in defining properties like limits and continuity. On a discrete set like integers one can't do these things in an interesting manner. $\endgroup$ – Paramanand Singh Apr 6 '20 at 8:16
0
$\begingroup$

You're right, we do not need to prove completeness of $\mathbb{R}$ in order to define the limit of a function. We define the limit of a function in such a way that we can get "arbitrarily" close to the suspected limit. The completeness of $\mathbb{R}$ comes into account when we ask "What values can the limit of our function be?"

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.