11.3 Continuity, Uniform Continuity, and Value Theorems

11.3 Continuity, Uniform Continuity, and Value Theorems

Continuity preserves the fundamental topological structures of metric spaces. On the GRE Mathematics Subject Test, questions test topological preimage formulations, the preservation of compactness and connectedness, and the distinction between pointwise, uniform, and Lipschitz continuity.

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Topological and Metric Definitions of Continuity

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces, and let $f: X \to Y$.

• Metric $\varepsilon$-$\delta$ Definition: $f$ is continuous at $x_0 \in X$ if for every $\varepsilon > 0$, there exists $\delta(\varepsilon, x_0) > 0$ such that $d_X(x, x_0) < \delta \implies d_Y(f(x), f(x_0)) < \varepsilon$.
• Topological Preimage Characterization: Global continuity is characterized without deltas: $f: X \\to Y \\text{ is continuous} \\iff f^{-1}(V) \\text{ is open in } X \\text{ for every open set } V \\subseteq Y.$ Equivalently, $f$ is continuous if and only if $f^{-1}(F)$ is closed in $X$ for every closed set $F \subseteq Y$.
• GRE Trap: Continuous functions do not preserve openness or closedness under forward images $f(U)$. For instance, $f(x) = x^2$ maps open $(-1, 1)$ to $[0, 1)$, which is not open.

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Preservation of Topological Invariants: EVT and IVT

Continuous functions preserve the topological invariants of compactness and connectedness:

Preservation of Compactness & Extreme Value Theorem

If $K \subseteq X$ is compact and $f: X \to Y$ is continuous, the image $f(K)$ is compact in $Y$.

• Extreme Value Theorem (EVT): If $K \subset \mathbb{R}^n$ is compact and $f: K \to \mathbb{R}$ is continuous, $f(K)$ is closed and bounded in $\mathbb{R}$. Thus $f$ attains an absolute maximum and minimum on $K$: $\\exists x_{\\min}, x_{\\max} \\in K \\quad \\text{such that} \\quad f(x_{\\min}) \\le f(x) \\le f(x_{\\max}) \\quad \\forall x \\in K.$
• Compact Homeomorphism Theorem: If $X$ is compact and $f: X \to Y$ is a continuous bijection onto a metric space $Y$, then $f^{-1}$ is continuous, so $f$ is a homeomorphism.

Preservation of Connectedness & Intermediate Value Theorem

If $E \subseteq X$ is connected and $f: X \to Y$ is continuous, the image $f(E)$ is connected in $Y$.

• Intermediate Value Theorem (IVT): If $f: [a, b] \to \mathbb{R}$ is continuous, $[a, b]$ is connected, so $f([a, b])$ is an interval. If $u$ lies between $f(a)$ and $f(b)$, there exists $c \in [a, b]$ with $f(c) = u$.

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Pointwise vs. Uniform Continuity

• Pointwise Continuity: $\forall x_0 \in X, \forall \varepsilon > 0, \exists \delta(\varepsilon, x_0) > 0$ such that $d_X(x, x_0) < \delta \implies d_Y(f(x), f(x_0)) < \varepsilon$.
• Uniform Continuity: $\forall \varepsilon > 0, \exists \delta(\varepsilon) > 0$ such that $\forall x, y \in X$: $d_X(x, y) < \\delta \\implies d_Y(f(x), f(y)) < \\varepsilon.$ A single $\delta$ works across the entire domain, independent of position.

Archetypal Non-Uniform Functions

1. $f(x) = 1/x$ on $(0, 1)$: Blows up as $x \to 0^+$; $\delta$ must shrink with $x^2$.
2. $f(x) = x^2$ on $\mathbb{R}$: Slope grows unboundedly as $x \to \infty$.
3. $f(x) = \sin(1/x)$ on $(0, 1)$: Bounded, but oscillates infinitely fast near $0$.
4. $f(x) = \sin(x^2)$ on $\mathbb{R}$: Bounded, but wave crests crowd closer together as $x \to \infty$.

Continuous Extension Theorem

A continuous function $f: (a, b) \to \mathbb{R}$ is uniformly continuous if and only if both one-sided limits $\lim_{x \to a^+} f(x)$ and $\lim_{x \to b^-} f(x)$ exist and are finite in $\mathbb{R}$.

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Heine-Cantor Theorem & Lipschitz Continuity

• Heine-Cantor Theorem: If $(X, d_X)$ is a compact metric space and $f: X \to Y$ is continuous, then $f$ is uniformly continuous. Any continuous function on $[a, b]$ is uniformly continuous.
• Lipschitz Continuity: $f$ is Lipschitz if $\exists L \ge 0$ such that $d_Y(f(x), f(y)) \le L \cdot d_X(x, y)$ for all $x, y$.
  • Choosing $\delta = \varepsilon / L$ confirms: Lipschitz $\implies$ Uniformly Continuous $\implies$ Continuous.
• Bounded Derivative Rule: If $f$ is differentiable on an interval and $|f'(x)| \le M$, then by the Mean Value Theorem, $|f(x) - f(y)| \le M|x - y|$, so $f$ is Lipschitz with $L = M$.

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Comparison Table: Continuity Hierarchy

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Step-by-Step Worked Problems

Problem 1: Uniform Continuity Verification on $[0, \infty)$

Show that $f(x) = \sqrt{x}$ is uniformly continuous on $[0, \infty)$.

Solution:

1. On $[0, 1]$, $f$ is continuous on a compact set, hence uniformly continuous by Heine-Cantor.
2. On $[1, \infty)$, $f'(x) = 1/(2\sqrt{x}) \le 1/2$. By MVT, $f$ is Lipschitz with $L = 1/2$.
3. Combining intervals: given $\varepsilon > 0$, choose $\delta = \min(1, \delta_{[0, 1]}, 2\varepsilon)$.
4. Alternatively, use the inequality $|\sqrt{x} - \sqrt{y}|^2 \le |x - y|$, which implies setting $\delta = \varepsilon^2$ works globally across $[0, \infty)$.

Problem 2: Topological Preimages of Level Sets

Determine whether $S = \{(x, y) \in \mathbb{R}^2 : 1 < x^2 + 2y^2 < 4\}$ is open.

Solution:

1. Let $g(x, y) = x^2 + 2y^2$. As a polynomial, $g: \mathbb{R}^2 \to \mathbb{R}$ is continuous.
2. $S = g^{-1}((1, 4))$.
3. Since $(1, 4)$ is an open interval in $\mathbb{R}$, its preimage $S$ under continuous $g$ is open in $\mathbb{R}^2$.

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GRE Exam Traps & Pitfalls

Trap 1: Assuming Bounded Functions Are Uniformly Continuous Boundedness does not guarantee uniform continuity. $f(x) = \sin(1/x)$ on $(0, 1)$ is bounded in $[-1, 1]$ but oscillates infinitely fast near $0$.

Trap 2: Forward Image Fallacy for Open Sets Continuity preserves open sets only under preimages $f^{-1}(V)$, not under forward images $f(U)$.

Trap 3: Inverting Continuous Bijections Without Compactness A continuous bijection $f: X \to Y$ is not a homeomorphism unless $X$ is compact. For example, $f: [0, 2\pi) \to S^1$ with $f(t) = e^{it}$ has a discontinuous inverse.