guide:56946c0de9

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Moving ahead with more theory, we would like to explain now an alternative formulation of the notion of continuity, which is quite abstract, and a bit difficult to understand and master when you are a beginner, but which is definitely worth learning, because it is quite powerful, solving some of the questions that we have left. Let us start with:

Definition

The open and closed sets are defined as follows:

• Open means that there is a small interval around each point.
• Closed means that our set is closed under taking limits.

As basic examples, the open intervals [math](a,b)[/math] are open, and the closed intervals [math][a,b][/math] are closed. Observe also that [math]\mathbb R[/math] itself is open and closed at the same time. Further examples, or rather results which are easy to establish, include the fact that the finite unions or intersections of open or closed sets are open or closed. We will be back to all this later, with some precise results in this sense. For the moment, we will only need:

Proposition

A set [math]O\subset\mathbb R[/math] is open precisely when its complement [math]C\subset\mathbb R[/math] is closed, and vice versa.

Show Proof

It is enough to prove the first assertion, since the “vice versa” part will follow from it, by taking complements. But this can be done as follows:

“[math]\implies[/math]” Assume that [math]O\subset\mathbb R[/math] is open, and let [math]C=\mathbb R-O[/math]. In order to prove that [math]C[/math] is closed, assume that [math]\{x_n\}_{n\in\mathbb N}\subset C[/math] converges to [math]x\in\mathbb R[/math]. We must prove that [math]x\in C[/math], and we will do this by contradiction. So, assume [math]x\notin C[/math]. Thus [math]x\in O[/math], and since [math]O[/math] is open we can find a small interval [math](x-\varepsilon,x+\varepsilon)\subset O[/math]. But since [math]x_n\to x[/math] this shows that [math]x_n\in O[/math] for [math]n[/math] big enough, which contradicts [math]x_n\in C[/math] for all [math]n[/math], and we are done.

“[math]\Longleftarrow[/math]” Assume that [math]C\subset\mathbb R[/math] is open, and let [math]O=\mathbb R-C[/math]. In order to prove that [math]O[/math] is open, let [math]x\in O[/math], and consider the intervals [math](x-1/n,x+1/n)[/math], with [math]n\in\mathbb N[/math]. If one of these intervals lies in [math]O[/math], we are done. Otherwise, this would mean that for any [math]n\in\mathbb N[/math] we have at least one point [math]x_n\in(x-1/n,x+1/n)[/math] satisfying [math]x_n\notin O[/math], and so [math]x_n\in C[/math]. But since [math]C[/math] is closed and [math]x_n\to x[/math], we get [math]x\in C[/math], and so [math]x\notin O[/math], contradiction, and we are done.

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As basic illustrations for the above result, [math]\mathbb R-(a,b)=(-\infty,a]\cup[b,\infty)[/math] is closed, and [math]\mathbb R-[a,b]=(-\infty,a)\cup(b,\infty)[/math] is open. Getting now back to functions, we have:

Theorem

A function is continuous precisely when [math]f^{-1}(O)[/math] is open, for any [math]O[/math] open. Equivalently, [math]f^{-1}(C)[/math] must be closed, for any [math]C[/math] closed.

Show Proof

Here the first assertion follows from definitions, and more specifically from the [math]\varepsilon,\delta[/math] definition of continuity, which was as follows:

[[math]] \forall x\in X,\forall\varepsilon \gt 0,\exists\delta \gt 0,|x-y| \lt \delta\implies|f(x)-f(y)| \lt \varepsilon [[/math]]

Indeed, if [math]f[/math] satisfies this condition, it is clear that if [math]O[/math] is open, then [math]f^{-1}(O)[/math] is open, and the converse holds too. As for the second assertion, this can be proved either directly, by using the [math]f(x_n)\to f(x)[/math] definition of continuity, or by taking complements.

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As a test for the above criterion, let us reprove the fact, that we know from Theorem 2.2, that if [math]f,g[/math] are continuous, so is [math]f\circ g[/math]. But this is clear, coming from:

[[math]] (f\circ g)^{-1}(O)=g^{-1}(f^{-1}(O)) [[/math]]

In short, not bad, because at least in relation with this specific problem, our proof using open sets is as simple as the simplest proof, namely the one using [math]f(x_n)\to f(x)[/math], and is simpler than the other proof that we know, namely the one with [math]\varepsilon,\delta[/math].

In order to reach to true applications of Theorem 2.9, we will need to know more about the open and closed sets. Let us begin with a useful result, as follows:

Proposition

The following happen:

• Union of open sets is open.
• Intersection of closed sets is closed.
• Finite intersection of open sets is open.
• Finite union of closed sets is closed.

Show Proof

Here (1) is clear from definitions, (3) is clear from definitions too, and (2,4) follow from (1,3) by taking complements [math]E\to E^c[/math], using the following formulae:

[[math]] \left(\bigcup_iE_i\right)^c=\bigcap_iE_i^c \quad,\quad \left(\bigcap_iE_i\right)^c=\bigcup_iE_i^c [[/math]]

Thus, we are led to the conclusions in the statement.

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As an important comment, (3,4) above do not hold when removing the finiteness assumption. Indeed, in what regards (3), the simplest counterexample here is:

[[math]] \bigcap_{n\in\mathbb N}\left(-\frac{1}{n}\,,\,\frac{1}{n}\right)=\{0\} [[/math]]

As for (4), here the simplest counterexample is as follows:

[[math]] \bigcup_{n\in\mathbb N}\left[0\,,\,1-\frac{1}{n}\right]=[0,1) [[/math]]

All this is quite interesting, and leads us to the question about what the open and closed sets really are. And fortunately, this question can be answered, as follows:

Theorem

The open and closed sets are as follows:

• The open sets are the disjoint unions of open intervals.
• The closed sets are the complements of these unions.

Show Proof

We have two assertions to be proved, the idea being as follows:

(1) We know that the open intervals are those of type [math](a,b)[/math] with [math]a \lt b[/math], with the values [math]a,b=\pm\infty[/math] allowed, and by Proposition 2.10 a union of such intervals is open.

(2) Conversely, given [math]O\subset\mathbb R[/math] open, we can cover each point [math]x\in O[/math] with an open interval [math]I_x\subset O[/math], and we have [math]O=\cup_xI_x[/math], so [math]O[/math] is a union of open intervals.

(3) In order to finish the proof of the first assertion, it remains to prove that the union [math]O=\cup_xI_x[/math] in (2) can be taken to be disjoint. For this purpose, our first observation is that, by approximating points [math]x\in O[/math] by rationals [math]y\in\mathbb Q\cap O[/math], we can make our union to be countable. But once our union is countable, we can start merging intervals, whenever they meet, and we are left in the end with a countable, disjoint union, as desired.

(4) Finally, the second assertion comes from Proposition 2.8.

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The above result is quite interesting, philosophically speaking, because contrary to what we have been doing so far, it makes the open sets appear quite different from the closed sets. Indeed, there is no way of having a simple description of the closed sets [math]C\subset\mathbb R[/math], similar to the above simple description of the open sets [math]O\subset\mathbb R[/math].

Moving towards more concrete things, and applications, let us formulate:

Definition

The compact and connected sets are defined as follows:

• Compact means that any open cover has a finite subcover.
• Connected means that it cannot be broken into two parts.

As basic examples, the closed bounded intervals [math][a,b][/math] are compact, as we know from the proof of Theorem 2.6, and so are the finite unions of such intervals. As for connected sets, the basic examples here are the various types of intervals, namely [math](a,b)[/math], [math](a,b][/math], [math][a,b)[/math], [math][a,b][/math], and it looks impossible to come up with more examples. In fact, we have:

Theorem

The compact and connected sets are as follows:

• The compact sets are those which are closed and bounded.
• The connected sets are the various types of intervals.

Show Proof

This is something quite intuitive, the idea being as follows:

(1) The fact that compact implies both closed and bounded is clear from our definition of compactness, because assuming non-closedness or non-boundedness leads to an open cover having no finite subcover. As for the converse, we know from the proof of Theorem 2.6 that any closed bounded interval [math][a,b][/math] is compact, and it follows that any [math]K\subset\mathbb R[/math] closed and bounded is a closed subset of a compact set, which follows to be compact.

(2) This is something which is obvious, and this regardless of what “cannot be broken into parts” in Definition 2.12 exactly means, mathematically speaking, with several possible definitions being possible here, all being equivalent. Indeed, [math]E\subset\mathbb R[/math] having this property is equivalent to [math]a,b\in E\implies[a,b]\subset E[/math], and this gives the result.

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We will be back to all this later in this book, when looking at open, closed, compact and connected sets in [math]\mathbb R^N[/math], or more general spaces, where things are more complicated than in [math]\mathbb R[/math]. Now with this discussed, let us go back to continuous functions. We have:

Theorem

Assuming that [math]f[/math] is continuous:

• If [math]K[/math] is compact, then [math]f(K)[/math] is compact.
• If [math]E[/math] is connected, then [math]f(E)[/math] is connected.

Show Proof

These assertions both follow from our definition of compactness and connectedness, as formulated in Definition 2.12. To be more precise:

(1) This comes from the fact that if a function [math]f[/math] is continuous, then the inverse function [math]f^{-1}[/math] returns an open cover into an open cover.

(2) This is something clear as well, because if [math]f(E)[/math] can be split into two parts, then by applying [math]f^{-1}[/math] we can split as well [math]E[/math] into two parts.

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Let us record as well the following useful generalization of Theorem 2.6:

Theorem

Any continuous function defined on a compact set

[[math]] f:X\to\mathbb R [[/math]]

is automatically uniformly continuous.

Show Proof

We can prove this exactly as Theorem 2.6, by using the compactness of [math]X[/math].

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You might perhaps ask at this point, were Theorems 2.14 and 2.15 worth all this excursion into open and closed sets. Good point, and here is our answer, a beautiful and powerful theorem based on the above, which can be used for a wide range of purposes:

Theorem

The following happen for a continuous function [math]f:[a,b]\to\mathbb R[/math]:

• [math]f[/math] takes all intermediate values between [math]f(a),f(b)[/math].
• [math]f[/math] has a minimum and maximum on [math][a,b][/math].
• If [math]f(a),f(b)[/math] have different signs, [math]f(x)=0[/math] has a solution.

Show Proof

All these statements are related, and are called altogether “intermediate value theorem”. Regarding now the proof, one way of viewing things is that since [math][a,b][/math] is compact and connected, the set [math]f([a,b])[/math] is compact and connected too, and so it is a certain closed bounded interval [math][c,d][/math], and this gives all the results. However, this is based on rather advanced technology, and it is possible to prove (1-3) directly as well.

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Along the same lines, we have as well the following result:

Theorem

Assuming that a function [math]f[/math] is continuous and invertible, this function must be monotone, and its inverse function [math]f^{-1}[/math] must be monotone and continuous too. Moreover, this statement holds both locally, and globally.

Show Proof

The fact that both [math]f[/math] and [math]f^{-1}[/math] are monotone follows from Theorem 2.16. Regarding now the continuity of [math]f^{-1}[/math], we want to prove that we have:

[[math]] x_n\to x\implies f^{-1}(x_n)\to f^{-1}(x) [[/math]]

But with [math]x_n=f(y_n)[/math] and [math]x=f(y)[/math], this condition becomes:

[[math]] f(y_n)\to f(y)\implies y_n\to y [[/math]]

And this latter condition being true since [math]f[/math] is monotone, we are done.

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As a basic application of Theorem 2.17, we have:

Proposition

The various usual inverse functions, such as the inverse trigonometric functions [math]\arcsin[/math], [math]\arccos[/math], [math]\arctan[/math], [math]{\rm arccot}[/math], are all continuous.

Show Proof

This follows indeed from Theorem 2.17, with a course the full discussion needing some explanations on bijectivity and domains. But you surely know all that, and in what concerns us, our claim is simply that these beasts are all continuous, proved.

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As another basic application of this, we have:

Proposition

The following happen:

• Any polynomial [math]P\in\mathbb R[X][/math] of odd degree has a root.
• Given [math]n\in2\mathbb N+1[/math], we can extract [math]\sqrt[n]{x}[/math], for any [math]x\in\mathbb R[/math].
• Given [math]n\in\mathbb N[/math], we can extract [math]\sqrt[n]{x}[/math], for any [math]x\in[0,\infty)[/math].

Show Proof

All these results come as applications of Theorem 2.16, as follows:

(1) This is clear from Theorem 2.16 (3), applied on [math][-\infty,\infty][/math].

(2) This follows from (1), by using the polynomial [math]P(z)=z^n-x[/math].

(3) This follows as well by applying Theorem 2.16 (3) to the polynomial [math]P(z)=z^n-x[/math], but this time on [math][0,\infty)[/math].

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There are many other things that can be said about roots of polyomials, and solutions of other equations of type [math]f(x)=0[/math], by using Theorem 2.16. We will be back to this.

As a concrete application, in relation with powers, we have the following result, completing our series of results regarding the basic mathematical functions:

Theorem

The function [math]x^a[/math] is defined and continuous on [math](0,\infty)[/math], for any [math]a\in\mathbb R[/math]. Moreover, when trying to extend it to [math]\mathbb R[/math], we have [math]4[/math] cases, as follows,

• For [math]a\in\mathbb Q_{odd}[/math], [math]a \gt 0[/math], the maximal domain is [math]\mathbb R[/math].
• For [math]a\in\mathbb Q_{odd}[/math], [math]a\leq0[/math], the maximal domain is [math]\mathbb R-\{0\}[/math].
• For [math]a\in\mathbb R-\mathbb Q[/math] or [math]a\in\mathbb Q_{even}[/math], [math]a \gt 0[/math], the maximal domain is [math][0,\infty)[/math].
• For [math]a\in\mathbb R-\mathbb Q[/math] or [math]a\in\mathbb Q_{even}[/math], [math]a\leq0[/math], the maximal domain is [math](0,\infty)[/math].

where [math]\mathbb Q_{odd}[/math] is the set of rationals [math]r=p/q[/math] with [math]q[/math] odd, and [math]\mathbb Q_{even}=\mathbb Q-\mathbb Q_{odd}[/math].

Show Proof

The idea is that we know how to extract roots by using Proposition 2.19, and all the rest follows by continuity. To be more precise:

(1) Assume [math]a=p/q[/math], with [math]p,q\in\mathbb N[/math], [math]p\neq0[/math] and [math]q[/math] odd. Given a number [math]x\in\mathbb R[/math], we can construct the power [math]x^a[/math] in the following way, by using Proposition 2.19:

[[math]] x^a=\sqrt[q]{x^p} [[/math]]

Then, it is straightforward to prove that [math]x^a[/math] is indeed continuous on [math]\mathbb R[/math].

(2) In the case [math]a=-p/q[/math], with [math]p,q\in\mathbb N[/math] and [math]q[/math] odd, the same discussion applies, with the only change coming from the fact that [math]x^a[/math] cannot be applied to [math]x=0[/math].

(3) Assume first [math]a\in\mathbb Q_{even}[/math], [math]a \gt 0[/math]. This means [math]a=p/q[/math] with [math]p,q\in\mathbb N[/math], [math]p\neq0[/math] and [math]q[/math] even, and as before in (1), we can set [math]x^a=\sqrt[q]{x^p}[/math] for [math]x\geq0[/math], by using Proposition 2.19. It is then straightforward to prove that [math]x^a[/math] is indeed continuous on [math][0,\infty)[/math], and not extendable either to the negatives. Thus, we are done with the case [math]a\in\mathbb Q_{even}[/math], [math]a \gt 0[/math], and the case left, namely [math]a\in\mathbb R-\mathbb Q[/math], [math]a \gt 0[/math], follows as well by continuity.

(4) In the cases [math]a\in\mathbb Q_{even}[/math], [math]a\leq0[/math] and [math]a\in\mathbb R-\mathbb Q[/math], [math]a\leq0[/math], the same discussion applies, with the only change coming from the fact that [math]x^a[/math] cannot be applied to [math]x=0[/math].

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Let us record as well a result about the function [math]a^x[/math], as follows:

Theorem

The function [math]a^x[/math] is as follows:

• For [math]a \gt 0[/math], this function is defined and continuous on [math]\mathbb R[/math].
• For [math]a=0[/math], this function is defined and continuous on [math](0,\infty)[/math].
• For [math]a \lt 0[/math], the domain of this function contains no interval.

Show Proof

This is a sort of reformulation of Theorem 2.20, by exchanging the variables, [math]x\leftrightarrow a[/math]. To be more precise, the situation is as follows:

(1) We know from Theorem 2.20 that things fine with [math]x^a[/math] for [math]x \gt 0[/math], no matter what [math]a\in\mathbb R[/math] is. But this means that things fine with [math]a^x[/math] for [math]a \gt 0[/math], no matter what [math]x\in\mathbb R[/math] is.

(2) This is something trivial, and we have of course [math]0^x=0[/math], for any [math]x \gt 0[/math]. As for the powers [math]0^x[/math] with [math]x\leq0[/math], these are impossible to define, for obvious reasons.

(3) Given [math]a \lt 0[/math], we know from Theorem 2.20 that we cannot define [math]a^x[/math] for [math]x\in\mathbb Q_{even}[/math]. But since [math]\mathbb Q_{even}[/math] is dense in [math]\mathbb R[/math], this gives the result.

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Summarizing, we have been quite successful with our theory of continuous functions, having how full results, regarding the definition and continuity property, for all basic functions from mathematics. All this is of course just a beginning, and we will be back to these functions on regular occasions, in what follows. In particular, we will discuss the function [math]a^x[/math] at the special value [math]a=e[/math], and its inverse [math]\log x[/math], at the end of this chapter.

General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].