What This Examination Tests
Connectedness arguments often combine several ideas, but each hypothesis has a distinct role. Connectedness rules out gaps and supports intermediate values; continuity transfers that information through a function; compactness can provide extrema or uniform bounds. This examination asks you to identify those roles and assemble them into proofs without treating the hypotheses as interchangeable.
The central new result concerns a continuous injective function on an interval. Such a function must preserve order throughout its domain: it is either strictly increasing or strictly decreasing. We will prove this using the Intermediate Value Theorem, then use strict monotonicity to prove that the inverse function is continuous. The examples also test two common errors: inferring connectedness of a domain from connectedness of its image, and assuming that continuity alone ensures injectivity.
Examination Result: Injectivity Forces Strict Monotonicity
A function is injective when distinct inputs always have distinct outputs. For an arbitrary function, that condition does not imply any consistent ordering of the outputs. Continuity on an interval changes the situation: if the function were to reverse its order, the Intermediate Value Theorem would force it to take some value twice.
Proof. If \(I\) has at most one point, the conclusion holds vacuously. Otherwise, choose \(a,b\in I\) with \(a<b\). Injectivity gives \(f(a)\ne f(b)\). We first establish that for every triple \(u<v<w\) in \(I\), the value \(f(v)\) lies strictly between \(f(u)\) and \(f(w)\).
Injectivity ensures that these three function values are distinct. Suppose, contrary to the claim, that \(f(v)\) is not between \(f(u)\) and \(f(w)\). Then it is either larger than both or smaller than both. Choose a real number \(t\) strictly between \(f(v)\) and the larger of \(f(u),f(w)\) in the first case, or strictly between the smaller of \(f(u),f(w)\) and \(f(v)\) in the second case. In either case, \(t\) lies strictly between \(f(u)\) and \(f(v)\), and also strictly between \(f(v)\) and \(f(w)\): when \(f(v)\) is above both endpoint values, \(t\) can be chosen between \(f(v)\) and \(\max\{f(u),f(w)\}\); when it is below both, choose \(t\) between \(f(v)\) and \(\min\{f(u),f(w)\}\). The Intermediate Value Theorem on \([u,v]\) and \([v,w]\) then gives points \(s\in(u,v)\) and \(r\in(v,w)\) with \(f(s)=t=f(r)\). Since \(s\ne r\), this contradicts injectivity. Thus \(f(v)\) is strictly between \(f(u)\) and \(f(w)\).
Suppose first that \(f(a)<f(b)\). We show that \(x<y\) implies \(f(x)<f(y)\). The triple property gives three useful comparisons. If \(x<b\), then \(f(x)<f(b)\): for \(x<a\), apply the property to \(x<a<b\); for \(a<x<b\), apply it to \(a<x<b\); and for \(x=a\), this is the chosen inequality. If \(y>a\), then \(f(y)>f(a)\), by applying the same property to \(a<b<y\) when \(y>b\), to \(a<y<b\) when \(a<y<b\), and using \(f(b)>f(a)\) when \(y=b\).
Now fix \(x<y\). If \(y\leq b\), then \(x<y<b\) whenever \(y<b\), and the triple property says \(f(y)\) lies between \(f(x)\) and \(f(b)\). Since \(f(x)<f(b)\), this implies \(f(x)<f(y)\). If \(y=b\), the same inequality follows from \(x<b\) and the comparison already established. If \(x\geq b\), then \(a<x<y\), and the triple property says \(f(x)\) lies between \(f(a)\) and \(f(y)\). Since \(f(x)>f(a)\), it follows that \(f(x)<f(y)\). In the remaining case \(x<b<y\), the comparisons give \(f(x)<f(b)<f(y)\). Hence \(f\) is strictly increasing.
If instead \(f(a)>f(b)\), apply the preceding argument to \(-f\), which is continuous and injective and satisfies \((-f)(a)<(-f)(b)\). It follows that \(-f\) is strictly increasing, so \(f\) is strictly decreasing. This proves the theorem. \(\square\)
The proof’s key move is the triple property. A failure of “the middle input has an output between the endpoint outputs” would create a repeated intermediate value on the two subintervals. Continuity supplies those intermediate values, and injectivity forbids the repetition. The interval hypothesis matters because it ensures the subintervals used in the argument lie in the domain.
Worked Example: Establishing Injectivity by Algebra
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^3+x\). For \(x<y\), factor the difference: \[ f(y)-f(x)=y^3+y-x^3-x=(y-x)(y^2+xy+x^2+1). \] The second factor is positive because \[ y^2+xy+x^2+1=(y+x/2)^2+3x^2/4+1>0. \] The first factor \(y-x\) is also positive. Therefore \(f(y)-f(x)>0\) whenever \(x<y\), so \(f\) is strictly increasing and, in particular, injective. The function is continuous, and its domain \(\mathbb{R}\) is an interval, so it also satisfies the theorem. Here the direct algebra verifies more than injectivity: it identifies which of the two possible monotonic directions occurs.
Worked Example: A Continuous Function That Is Not Injective
Consider \(g:[-2,2]\to\mathbb{R}\) defined by \(g(x)=x^2\). It is continuous, but \(g(-1)=1=g(1)\), even though \(-1\ne1\). Thus it is not injective and, as expected, is neither strictly increasing nor strictly decreasing on the whole interval: \(g(-2)=4>g(-1)=1\), whereas \(g(1)=1<g(2)=4\). Continuity alone does not imply monotonicity. The theorem’s injectivity hypothesis is essential.
Examination Result: The Inverse Is Continuous
Strict monotonicity makes a function one-to-one, so an inverse function exists from its range back to its domain. The following result shows that when the original function is continuous on an interval, its inverse is continuous too. This is a useful conclusion about the structure of the function, not merely about the connectedness of its range.
Proof. A strictly monotone function is injective, so \(f^{-1}\) is well-defined on \(f(I)\). Fix \(y_0\in f(I)\), and write \(x_0=f^{-1}(y_0)\). We prove continuity at \(y_0\). Let \(\varepsilon>0\). We will find \(\delta>0\) such that for every \(y\in f(I)\), if \(|y-y_0|<\delta\), then \(|f^{-1}(y)-x_0|<\varepsilon\).
First suppose \(f\) is strictly increasing. If there is a point \(u\in I\) with \(u\leq x_0-\varepsilon\), define \(t_-=x_0-\varepsilon/2\). Because \(u\leq x_0-\varepsilon<t_-<x_0\) and \(I\) is an interval, \(t_-\in I\). Also \(t_->x_0-\varepsilon\), so any \(x\in I\) satisfying \(x\leq x_0-\varepsilon\) has \(x<t_-\). Strict increase gives \(f(x)<f(t_-)<f(x_0)=y_0\). Thus such an \(x\) cannot have \(f(x)\) sufficiently close to \(y_0\): specifically, if \(|f(x)-y_0|<y_0-f(t_-)\), then \(f(x)>f(t_-)\), a contradiction. If there is no such \(u\), no point of \(I\) lies at or to the left of \(x_0-\varepsilon\), so the left side requires no restriction.
Likewise, if there is a point \(v\in I\) with \(v\geq x_0+\varepsilon\), set \(t_+=x_0+\varepsilon/2\). Then \(x_0<t_+<x_0+\varepsilon\leq v\), \(t_+\in I\), and \(t_+<x_0+\varepsilon\). For \(x\in I\) with \(x\geq x_0+\varepsilon\), strict increase gives \(f(x)>f(t_+)>y_0\). Such \(x\) is excluded whenever \(|f(x)-y_0|<f(t_+)-y_0\).
For each side on which a point \(u\) or \(v\) exists, take the corresponding positive bound \(y_0-f(t_-)\) or \(f(t_+)-y_0\). Let \(\delta\) be the minimum of the bounds that exist; if neither side has such a point, choose \(\delta=1\). In every case \(\delta>0\). If \(y\in f(I)\) and \(|y-y_0|<\delta\), its preimage cannot be at distance at least \(\varepsilon\) from \(x_0\) on either side, so \(|f^{-1}(y)-x_0|<\varepsilon\). This proves continuity at \(y_0\) when \(f\) is increasing. If \(f\) is strictly decreasing, apply the increasing case to \(-f\); the inverse of \(-f\) is \(y\mapsto f^{-1}(-y)\), and the resulting continuity statement gives continuity of \(f^{-1}\) at \(y_0\) as well. Since \(y_0\) was arbitrary, \(f^{-1}\) is continuous on \(f(I)\). \(\square\)
The proof also accounts for endpoints and one-sided domains. If there are no domain points at least \(\varepsilon\) away on one side of \(x_0\), that side cannot cause a failure of continuity and needs no bound. When such points do exist, the interval property supplies an intermediate domain point whose function value gives a positive exclusion margin.
Worked Example: The Inverse of the Cube Function
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^3\). If \(x<y\), then \[ y^3-x^3=(y-x)(y^2+xy+x^2)>0. \] Indeed, \(y-x>0\), and \(y^2+xy+x^2=(y+x/2)^2+3x^2/4\) can equal zero only if \(x=y=0\), which is incompatible with \(x<y\). Thus \(f\) is strictly increasing. It is continuous, and its image is \(\mathbb{R}\): for each real \(y\), there is a real cube root \(x=\sqrt[3]{y}\) with \(f(x)=y\). The theorem therefore gives a continuous inverse \(f^{-1}(y)=\sqrt[3]{y}\) on \(\mathbb{R}\). In particular, the inverse is continuous at \(0\), despite its inverse-function formula not having a finite derivative there.
Image Connectedness Does Not Determine the Domain
A continuous image of a connected set is connected, as established by the Continuous Images of Connected Sets Theorem. It is tempting to reverse that implication, but it is false: a disconnected domain can have a connected image. The next example makes the distinction explicit and checks the image rather than relying on a sketch.
Worked Example: A Disconnected Domain with an Interval as Its Image
Set \(E=[-2,-1]\cup[1,2]\) and define \(h:E\to\mathbb{R}\) by \(h(x)=x^2\). The set \(E\) is disconnected: its two displayed intervals are nonempty, disjoint, and separated by a gap. On the first interval, \(x\in[-2,-1]\) implies \(x^2\in[1,4]\), and every \(t\in[1,4]\) is attained there by \(x=-\sqrt{t}\). On the second interval, \(x\in[1,2]\) also gives \(x^2\in[1,4]\), and every \(t\in[1,4]\) is attained by \(x=\sqrt{t}\). Hence \(h(E)=[1,4]\), which is connected. There is no contradiction with the continuous-image theorem: that theorem says a connected domain has a connected image, not that every domain with connected image is connected.
How to Audit a Connectedness Proof
In an examination proof, the most efficient first step is to name the job of each assumption. A continuity hypothesis may justify the Intermediate Value Theorem or the openness of a preimage. A connectedness hypothesis may prevent the image from having a gap. A compactness hypothesis may provide an extremum, but compactness alone does not fill gaps. Checking these roles often identifies a missing hypothesis before lengthy calculations begin.
Determine whether the claim concerns the domain, the image, a level set, or the range of an inverse.
Use connectedness for interval structure and continuity for intermediate values; use compactness only when an extremum or compactness conclusion is needed.
In monotonicity arguments, handle points on either side of the chosen anchor pair, including when a point equals an anchor.
A theorem about images of connected sets does not automatically provide a converse about domains.
A final useful distinction is between “connected” and “path connected.” In \(\mathbb{R}\), connected subsets are intervals and are path connected, but connectedness and path connectedness do not agree in every metric space; the Topologist’s Sine Curve discussed earlier in the course demonstrates this. When a problem is explicitly about a subset of the real line, use the available interval characterization. In a more general space, do not import that characterization without justification.
Check Your Understanding
Use the stated hypotheses carefully; when a proof fails, identify which step is unsupported.
- Why does the proof of strict monotonicity require the domain to contain the subinterval between any two of its points?
- In the triple-property argument, how does a value between \(f(v)\) and both endpoint values contradict injectivity?
- For a strictly decreasing continuous function on an interval, which function can be used to reduce the inverse-continuity proof to the increasing case?
- Give the exact image of \(x\mapsto x^2\) on \([-2,-1]\cup[1,2]\), and explain why this does not imply that the domain is connected.
- Which hypothesis in the strict-monotonicity theorem rules out repeated outputs, and which hypothesis makes the Intermediate Value Theorem available?