From Connectedness to Intermediate Values
The Intermediate Value Theorem says that a continuous real-valued function cannot pass from one value to another while skipping a value in between. Connectedness gives a structural explanation: an interval is connected, continuous maps preserve connectedness, and a connected subset of the real line is an interval. Thus the function’s image cannot have a gap.
All real intervals below carry the subspace topology inherited from \(\mathbb{R}\) with its standard topology. A map from an interval into \(\mathbb{R}\) is continuous with respect to that subspace topology and the standard topology on \(\mathbb{R}\).
Two results from the earlier material on connectedness supply the key steps: every interval in \(\mathbb{R}\) is connected, and a connected subset of \(\mathbb{R}\) is an interval. We also use the result that continuous images preserve connectedness: if \(X\) is connected and \(f:X\to Y\) is continuous, then \(f(X)\), with the subspace topology from \(Y\), is connected. These results apply to arbitrary topological spaces and continuous maps; here we use them with an interval as the domain and \(\mathbb{R}\) as the codomain.
The Image of an Interval Is an Interval
Proof. If \(J=\varnothing\), then \(f(J)=\varnothing\), which is an interval. Otherwise, \(J\) is connected because every interval in \(\mathbb{R}\) is connected. Since \(f\) is continuous, the continuous-image theorem gives that \(f(J)\) is connected. The theorem that connected subsets of \(\mathbb{R}\) are intervals now shows that \(f(J)\) is an interval. \(\square\)
This proposition is the connectedness form of the Intermediate Value Theorem. It asserts something about the whole range, not just about a chosen pair of inputs: whenever two values belong to the range, every real number between those values belongs to the range as well.
Proof. The interval \([a,b]\) is connected. By continuity, \(f([a,b])\) is connected, and therefore it is an interval in \(\mathbb{R}\). Both \(f(a)\) and \(f(b)\) belong to \(f([a,b])\). Since \(y\) lies between them, the definition of interval implies \(y\in f([a,b])\). By the definition of image, there is some \(c\in[a,b]\) with \(f(c)=y\). If \(y\) is strictly between the endpoint values, then \(y\ne f(a)\) and \(y\ne f(b)\); consequently \(c\ne a\) and \(c\ne b\), so \(c\in(a,b)\). \(\square\)
The strict inequalities are useful when an interior point is required. When \(y\) equals an endpoint value, the theorem only guarantees a preimage in the closed interval; that preimage may be the endpoint itself. The theorem also does not claim that \(c\) is unique. Uniqueness requires additional information about \(f\), such as strict monotonicity, and is not a consequence of connectedness alone.
Worked Applications
Worked Example: A Sign Change Gives a Root
Show that \(f(x)=x^3-3x+1\) has a zero in \((0,1)\).
The domain \([0,1]\) is an interval, and \(f\) is a polynomial, hence continuous on \([0,1]\). Evaluate the endpoint values directly:
The target value \(0\) satisfies \(-1<0<1\), so it lies strictly between \(f(1)=-1\) and \(f(0)=1\). The Intermediate Value Theorem therefore gives a point \(c\in(0,1)\) such that \(f(c)=0\).
Check: The endpoint values have opposite signs, as the calculation shows: \(f(0)=1>0\) and \(f(1)=-1<0\). The conclusion is existence of a root in the open interval; this argument does not calculate its exact value or prove that it is the only root there.
Worked Example: A Value Between Two Outputs
Let \(f(x)=2x+3\) on \([-1,2]\). Use the theorem to locate an input where the function takes the value \(4\), and then check the input exactly.
The function is linear, hence continuous, and the domain is an interval. Its endpoint values are
Since \(1<4<7\), the theorem guarantees some \(c\in(-1,2)\) with \(f(c)=4\). To find and check such an input, solve the equation:
Now \(-1<\frac{1}{2}<2\), and substituting gives \(f(\frac{1}{2})=2(\frac{1}{2})+3=1+3=4\). Thus \(c=\frac{1}{2}\) is a preimage of the target value, in agreement with the theorem.
Worked Example: Determining a Range on an Interval
Find the range of \(f(x)=x^2\) on \([0,2]\), using the Intermediate Value Theorem to establish that no values between the extremes are omitted.
For every \(x\in[0,2]\), the inequalities \(0\le x\le2\) imply \(0\le x^2\le4\). Thus \(f([0,2])\subseteq[0,4]\). The endpoint calculations give \(f(0)=0^2=0\) and \(f(2)=2^2=4\). By the theorem, every \(y\) with \(0\le y\le4\) is attained: it lies between the endpoint values, so some \(c\in[0,2]\) satisfies \(c^2=y\). Hence \([0,4]\subseteq f([0,2])\), and the two inclusions give
Check: Every output is at least \(0\) and at most \(4\) by the square bounds above. Conversely, the connected-image argument establishes that every value in that entire closed interval occurs; the endpoint values \(0\) and \(4\) themselves occur at \(x=0\) and \(x=2\).
Why the Domain Hypothesis Matters
Continuity alone does not guarantee intermediate values if the domain is disconnected. For example, set \(A=[0,1]\cup[2,3]\) with its subspace topology from \(\mathbb{R}\), and let \(f:A\to\mathbb{R}\) be \(f(x)=x\). This map is continuous, but \(f(A)=A\), which misses \(1.5\), even though \(1.5\) lies strictly between \(f(0)=0\) and \(f(3)=3\).
The failure is consistent with the connectedness argument: \(A\) is not an interval, and it is disconnected. In fact, \(A\cap(-1,1.5)=[0,1]\) and \(A\cap(1.5,4)=[2,3]\) are disjoint, nonempty, relatively open subsets of \(A\) whose union is \(A\). Thus the continuous-image theorem does not give connectedness of \(f(A)\). Applying the Intermediate Value Theorem here would be a mistake because its domain hypothesis is absent.
A second point to keep in view is the distinction between the theorem’s conclusion and a method for explicitly solving an equation. Connectedness proves that a preimage exists; it does not generally provide a formula for that preimage, tell how many preimages there are, or show that a numerical approximation is accurate. Those are separate questions. The theorem’s strength is that it establishes existence without requiring a formula for the input.
| Ingredient | Role in the argument |
|---|---|
| The domain is an interval | It is connected. |
| The function is continuous | Its image is connected. |
| The codomain is \(\mathbb{R}\) | Every connected subset is an interval. |
| The target lies between two outputs | The interval property puts the target in the image. |
Check Your Understanding
For each question, identify the connectedness or interval fact that justifies your conclusion.
- Suppose \(f:[-2,3]\to\mathbb{R}\) is continuous, \(f(-2)=5\), and \(f(3)=-1\). What does the Intermediate Value Theorem guarantee about the value \(2\)?
- Why does continuity of a function on \([0,1]\cup[2,3]\) alone not guarantee that it takes every value between its values at \(0\) and \(3\)? Give an explicit continuous function illustrating the failure.
- If \(f:[a,b]\to\mathbb{R}\) is continuous and \(f(a)=f(b)\), what does the theorem guarantee for the target \(y=f(a)\)? Does it guarantee a preimage in \((a,b)\)?
- State which hypotheses are needed to conclude that the image of a domain is an interval, and explain the role of each.
- For \(g(x)=x^2-1\) on \([0,2]\), use endpoint values to show that \(g(x)=2\) has a solution in \((0,2)\). Then check an exact solution by substitution.