Level Sets

Equations Define Sets

An equation such as $F(x)=c$ can be viewed not only as a condition to solve, but also as a way to define a subset of the domain of $F$. The resulting set is called a level set. The Implicit Function Theorem, discussed in the previous tutorial, explains when such a set has a particularly simple local description: near a suitable point, some coordinates can be expressed as functions of the others.

This local description requires care. A full-rank derivative provides a powerful sufficient condition for a level set to be a graph, but it is not a necessary condition. Level sets also have useful global properties that follow from continuity, without any rank assumption.

Definition: Let $U\subseteq\mathbb{R}^n$, let $F:U\to\mathbb{R}^k$, and let $c\in\mathbb{R}^k$. The level set of $F$ at $c$ is $$ F^{-1}(\{c\})=\{x\in U:F(x)=c\}. $$ When $k=1$, this is a scalar level set; for example, $f(x)=c$. The zero set is the special case $c=0$.

A level set may be empty, may consist of isolated points, or may contain curves or higher-dimensional pieces. Its appearance depends on both the function and the chosen level. Replacing $F(x)=c$ with $(F-c)(x)=0$ gives the same set, so the zero-level language can always be used after shifting the function by $c$.

Worked Example: A Sphere as a Level Set

Define $F:\mathbb{R}^3\to\mathbb{R}$ by $F(x,y,z)=x^2+y^2+z^2$. The level set at $1$ consists of exactly those points satisfying $x^2+y^2+z^2=1$, which is the unit sphere. For instance, $(0,0,1)$ belongs to it because $0^2+0^2+1^2=1$, whereas $(1,1,0)$ does not because $1^2+1^2+0^2=2\neq1$.

The derivative is the row matrix $DF(x,y,z)=(2x,2y,2z)$. At $(0,0,1)$, its $z$-derivative is $2$, which is nonzero. The Implicit Function Theorem therefore allows $z$ to be solved locally as a function of $x$ and $y$. On the part of the sphere near $(0,0,1)$, the formula is $z=\sqrt{1-x^2-y^2}$. Substituting this expression into the equation gives $x^2+y^2+(\sqrt{1-x^2-y^2})^2=1$, whenever $x^2+y^2<1$. The positive square root selects the portion near the north pole; the negative square root describes the portion near the south pole.

Continuity and Closed Level Sets

The local graph theorem concerns the derivative near a particular point. A simpler global fact comes from continuity alone: the preimage of a closed set under a continuous function is closed in the domain. Since a single point $\{c\}$ is closed in $\mathbb{R}^k$, this applies directly to level sets. Here “closed in $U$” means closed relative to the domain $U$; the level set need not be closed in all of $\mathbb{R}^n$ if $U$ itself is not closed.

Theorem (Closed Level Sets): If $U\subseteq\mathbb{R}^n$, $F:U\to\mathbb{R}^k$ is continuous, and $c\in\mathbb{R}^k$, then $F^{-1}(\{c\})$ is closed relative to $U$. In particular, if $U=\mathbb{R}^n$, the level set is closed in $\mathbb{R}^n$.

Proof. The singleton $\{c\}$ is closed in $\mathbb{R}^k$. Continuity of $F$ implies that the preimage of every closed subset of $\mathbb{R}^k$ is closed relative to $U$. Applying this fact to $\{c\}$ gives that $F^{-1}(\{c\})$ is closed relative to $U$, as claimed. When $U=\mathbb{R}^n$, relative closedness is simply closedness in $\mathbb{R}^n$. $\square$

Worked Example: A Hyperbola with Two Components

Let $F:\mathbb{R}^2\to\mathbb{R}$ be $F(x,y)=xy$. Its level set at $1$ is $\{(x,y):xy=1\}$. Because $F$ is continuous, the Closed Level Sets Theorem says this set is closed in $\mathbb{R}^2$. No point of the set has $x=0$, since $0\cdot y=0$, not $1$. Thus the equation can be solved globally for $y$ as $y=1/x$ on the domain $x\neq0$. It has one part where $x>0,y>0$ and another where $x<0,y<0$.

The derivative is $DF(x,y)=(y,x)$. At every point of the level set, $xy=1$, so neither coordinate is zero and $DF(x,y)\neq(0,0)$. In particular, the derivative is nonzero at $(1,1)$, where $1\cdot1=1$, and the local graph conclusion applies there. The global formula $y=1/x$ is available in this example, but the local theorem by itself promises only a description near the chosen point; it does not establish a single global graph description for every level set.

Regular Points and Local Graphs

For a map $F:U\to\mathbb{R}^k$, its derivative at $a$ is a linear map from $\mathbb{R}^n$ to $\mathbb{R}^k$. When this derivative has rank $k$, it is onto. Equivalently, the $k$-by-$n$ Jacobian matrix has $k$ linearly independent rows. Such a point is called regular for $F$. A value $c$ is called a regular value if every point $a$ satisfying $F(a)=c$ is regular; if the level set is empty, this condition holds vacuously.

Definition: Let $F:U\to\mathbb{R}^k$ be continuously differentiable, where $U\subseteq\mathbb{R}^n$ is open. A point $a\in U$ is a regular point of $F$ if $DF(a)$ has rank $k$. A value $c\in\mathbb{R}^k$ is a regular value if every point in $F^{-1}(\{c\})$ is a regular point.

At a regular point, some $k$ columns of the Jacobian form an invertible $k$-by-$k$ matrix. Reordering coordinates if needed, write a point of $\mathbb{R}^n$ as $(x,y)$, where $x\in\mathbb{R}^{n-k}$ and $y\in\mathbb{R}^k$, so that $D_yF(a)$ is invertible. The Implicit Function Theorem then solves $F(x,y)=c$ for $y$ as a function of $x$, locally. The next result makes that conclusion precise.

Theorem (Local Graph Theorem for a Regular Level Set): Let $U\subseteq\mathbb{R}^{n-k}\times\mathbb{R}^k$ be open, let $F:U\to\mathbb{R}^k$ be continuously differentiable, and suppose $F(a,b)=c$. If the $k$-by-$k$ matrix $D_yF(a,b)$ is invertible, then there are neighborhoods $X$ of $a$ and $Y$ of $b$, and a continuously differentiable function $\phi:X\to Y$, such that $$ F(x,y)=c \quad\Longleftrightarrow\quad y=\phi(x) $$ for $(x,y)\in X\times Y$. Consequently, in this neighborhood the level set is the graph $\{(x,\phi(x)):x\in X\}$.

Proof. Define $G:U\to\mathbb{R}^k$ by $G(x,y)=F(x,y)-c$. Then $G(a,b)=0$, and $D_yG(a,b)=D_yF(a,b)$ is invertible. The Implicit Function Theorem gives an open neighborhood $W\subseteq U$ of $(a,b)$, an open neighborhood $X_0$ of $a$, and a continuously differentiable function $\phi:X_0\to\mathbb{R}^k$ whose graph lies in $W$, with $G(x,\phi(x))=0$, and such that any $(x,y)\in W$ satisfying $G(x,y)=0$ has $y=\phi(x)$. Choose open neighborhoods $X_1$ of $a$ and $Y$ of $b$ with $X_1\times Y\subseteq W$. Since $\phi(a)=b$, shrink $X_1\cap X_0$ to an open neighborhood $X$ of $a$ so that $\phi(X)\subseteq Y$. Thus, for $x\in X$ and $y\in Y$, $G(x,y)=0$ if and only if $y=\phi(x)$. Since $G(x,y)=0$ is equivalent to $F(x,y)=c$, this is the stated equivalence. Every point of the level set in $X\times Y$ therefore has the form $(x,\phi(x))$, and every such point belongs to the level set. $\square$

The rank-$k$ condition gives a way to find the required invertible block: choose $k$ independent columns of $DF(a,b)$, and use those coordinates as $y$. This is why full rank guarantees a local graph over the remaining $n-k$ coordinates. It does not say that every level set that happens to be a graph must satisfy the rank condition.

Worked Example: A Line Defined by Two Equations

Consider $F:\mathbb{R}^3\to\mathbb{R}^2$ defined by $F(x,y,z)=(x+y,\ y-z)$. The level set at $(0,0)$ consists of solutions to $x+y=0$ and $y-z=0$. Let $t=y$. The first equation gives $x=-t$, and the second gives $z=t$. Therefore the full solution set is $\{(-t,t,t):t\in\mathbb{R}\}$, a line in $\mathbb{R}^3$.

The Jacobian matrix is \[ DF(x,y,z)= \begin{pmatrix} 1&1&0\\ 0&1&-1 \end{pmatrix}. \] Its first and third columns form the matrix $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$, whose determinant is $-1$, so the rank is $2$. The theorem lets us solve for $(x,z)$ in terms of $y$: $x=-y$ and $z=y$. Substitution checks both equations at every point on the stated line: $(-t)+t=0$ and $t-t=0$. The two independent equations give a one-dimensional local graph here, not a plane.

Rank Is Sufficient, Not Necessary

The condition in the Local Graph Theorem is a hypothesis that allows the Implicit Function Theorem to produce a graph. It should not be reversed into a claim that full rank is necessary whenever a level set is a graph. A level set can have a simple graph description even at points where the derivative loses rank. Conversely, a singular level can behave quite differently from a regular one.

Worked Example: A Graph at a Singular Level

Let $F:\mathbb{R}^2\to\mathbb{R}$ be $F(x,y)=x^2$, and consider the level $c=0$. The equation $x^2=0$ holds exactly when $x=0$, so the level set is the line $\{(0,y):y\in\mathbb{R}\}$. This is a graph: it can be written as $x=\phi(y)$, where $\phi(y)=0$ for every $y$.

However, $DF(x,y)=(2x,0)$, and hence $DF(0,y)=(0,0)$ at every point of this level set. Its rank is $0$, not $1$. Thus the local graph description exists despite the failure of the rank condition. The condition is sufficient for the graph conclusion via the Implicit Function Theorem, but it is not necessary for a level set to be a graph.

A singular point can also occur on a level set that does not resemble a regular curve or surface. For example, $F(x,y)=x^2+y^2$ has level $0$ equal to the single point $(0,0)$, since a sum of two squares is zero only if both squares are zero. Its derivative at the origin is the zero map. This contrasts with the line in the previous example: the same failure of rank does not determine the shape of the level set. Additional analysis is needed at singular points.

Using Level Sets Carefully

The level-set viewpoint separates two kinds of information. Continuity gives a closedness conclusion for the whole level set relative to the domain. Differentiability and a full-rank condition give a local graph description near a specified point. These are different conclusions with different hypotheses: closedness does not imply a local graph, and the local graph theorem does not describe the entire level set.

In applications, first check the equation $F(a)=c$ by substituting the proposed point. Then compute the derivative and check which coordinates, if any, form an invertible block. If the relevant block is invertible, the Implicit Function Theorem identifies coordinates that can be solved locally. If the rank is deficient, do not conclude that a graph is impossible; instead, examine the equations directly, as in the $x^2=0$ example. Likewise, a local graph near one point does not rule out other components or distant parts of the level set, as the two-component hyperbola illustrates.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What is the level set of a map $F:U\to\mathbb{R}^k$ at a value $c$?
Why is a level set of a continuous map closed relative to its domain?
What role does an invertible $k$-by-$k$ derivative block play in the local graph theorem?
Why does the set $x^2=0$ show that full rank is not necessary for a level set to be a graph?
Does a local graph description near one point determine the global shape or number of components of the full level set? Explain.

What You'll Learn