Continuous Image of Connected and Compact Spaces

Two of the most powerful theorems in topology state that continuous images preserve both connectedness and compactness. These results unify many classical theorems from analysis, including the intermediate value theorem and the extreme value theorem.

Continuous Image Preserves Connectedness

Theorem2.8Continuous Image of a Connected Space

Let $f: X \to Y$ be a continuous map and let $X$ be connected. Then $f(X)$ is connected (as a subspace of $Y$ ).

Proof

Suppose for contradiction that $f(X)$ is disconnected. Then there exist nonempty open sets $U, V$ in $f(X)$ with $f(X) = U \cup V$ and $U \cap V = \emptyset$ .

Since $f$ is continuous and the inclusion $\iota: f(X) \hookrightarrow Y$ gives the subspace topology, $f^{-1}(U)$ and $f^{-1}(V)$ are open in $X$ . They are nonempty (since $U$ and $V$ are nonempty subsets of $f(X)$ ), disjoint (since $f^{-1}(U) \cap f^{-1}(V) = f^{-1}(U \cap V) = f^{-1}(\emptyset) = \emptyset$ ), and cover $X$ (since $f^{-1}(U) \cup f^{-1}(V) = f^{-1}(U \cup V) = f^{-1}(f(X)) = X$ ).

This contradicts the connectedness of $X$ .

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Continuous Image Preserves Compactness

Theorem2.9Continuous Image of a Compact Space

Let $f: X \to Y$ be a continuous map and let $X$ be compact. Then $f(X)$ is compact.

Proof

Let $\{V_\alpha\}_{\alpha \in A}$ be an open cover of $f(X)$ (open in the subspace topology on $f(X)$ , or equivalently, open sets in $Y$ that cover $f(X)$ ).

Then $\{f^{-1}(V_\alpha)\}_{\alpha \in A}$ is an open cover of $X$ (each $f^{-1}(V_\alpha)$ is open since $f$ is continuous, and they cover $X$ since $f(X) \subseteq \bigcup V_\alpha$ ).

By compactness of $X$ , there is a finite subcover $\{f^{-1}(V_{\alpha_1}), \ldots, f^{-1}(V_{\alpha_n})\}$ . Then: $f(X) \subseteq f\left(\bigcup_{i=1}^n f^{-1}(V_{\alpha_i})\right) \subseteq \bigcup_{i=1}^n V_{\alpha_i}.$

So $\{V_{\alpha_1}, \ldots, V_{\alpha_n}\}$ is a finite subcover of $f(X)$ .

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Classical Consequences

Theorem2.10Extreme Value Theorem

Let $X$ be a compact topological space and $f: X \to \mathbb{R}$ a continuous function. Then $f$ attains its maximum and minimum values. That is, there exist $x_{\min}, x_{\max} \in X$ such that: $f(x_{\min}) \leq f(x) \leq f(x_{\max}) \quad \text{for all } x \in X.$

Proof

By Theorem 2.9, $f(X)$ is compact in $\mathbb{R}$ . By the Heine--Borel theorem, $f(X)$ is closed and bounded. Since $f(X)$ is bounded, $\sup f(X)$ and $\inf f(X)$ exist. Since $f(X)$ is closed, these values belong to $f(X)$ .

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ExampleIntermediate Value Theorem from Connectedness

The classical intermediate value theorem follows immediately: if $f: [a, b] \to \mathbb{R}$ is continuous and $c$ is between $f(a)$ and $f(b)$ , then $c \in f([a,b])$ .

Since $[a,b]$ is connected and $f$ is continuous, $f([a,b])$ is a connected subset of $\mathbb{R}$ , hence an interval. Since $f(a)$ and $f(b)$ belong to this interval, every value between them does as well.

Further Applications

Theorem2.11Continuous Bijection from Compact to Hausdorff

Let $f: X \to Y$ be a continuous bijection where $X$ is compact and $Y$ is Hausdorff. Then $f$ is a homeomorphism.

Proof

We show $f$ is a closed map. Let $C \subseteq X$ be closed. Since $X$ is compact, $C$ is compact (closed subset of a compact space). By Theorem 2.9, $f(C)$ is compact. Since $Y$ is Hausdorff, $f(C)$ is closed in $Y$ .

So $f$ is a continuous closed bijection, hence a homeomorphism.

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ExampleHomeomorphism via Compact-Hausdorff

The quotient map $q: [0, 1] \to S^1$ defined by $q(t) = e^{2\pi i t}$ identifies $0 \sim 1$ . The induced map $\bar{q}: [0,1]/{\sim} \to S^1$ is a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism.

This is a standard technique: to show that a quotient space is homeomorphic to a known space, construct a continuous bijection from the quotient to the target and apply Theorem 2.11.

RemarkTopological Properties Preserved by Continuous Images

Beyond connectedness and compactness, continuous images also preserve:

Path-connectedness: If $X$ is path-connected and $f: X \to Y$ is continuous, then $f(X)$ is path-connected (compose paths with $f$ ).
$\sigma$ -compactness: A countable union of compact sets maps to a countable union of compact sets.
Separability: The continuous image of a separable space is separable.
Lindel"of property: Continuous images of Lindel"of spaces are Lindel"of.

However, continuous images do not generally preserve: Hausdorff, regular, normal, first-countable, second-countable, or metrizable.