TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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The image of a compact subset of $\mathbb{R}$ under a continuous real function is compact

Statement

Let $A \subseteq \mathbb{R}$ , let $f : A \to \mathbb{R}$ be continuous on $A$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point) and let $K \subseteq A$ be compact in the sense of Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset, that is, every family of open subsets of $\mathbb{R}$ whose union contains $K$ has a finite subfamily whose union contains $K$ . Then the image

$f[K] \;:=\; \{\, f(x) : x \in K \,\}$

is a compact subset of $\mathbb{R}$ .

This theorem is stated twice in this library, on purpose. Its metric-space twin is The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset, proved from the cover machinery of metric spaces; the proof below is $\mathbb{R}$ -native and uses only open subsets of $\mathbb{R}$ and the definition of continuity of this page. The duplication is deliberate and is acknowledged in exactly one place, Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$ , continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace later on this page, which proves that the two notions of "compact subset of $\mathbb{R}$ " and the two notions of "continuous" coincide, so that the two theorems really are the same statement in two vocabularies.

No choice principle is used. The covering family built below is cut out by a property rather than selected point by point, and the only selection made is from a finite list, which needs no axiom.

Facts & Assumptions

Given: A set $A \subseteq \mathbb{R}$ , a function $f : A \to \mathbb{R}$ continuous on $A$ , and a compact set $K \subseteq A$ ; $f[K] = \{\, f(x) : x \in K \,\}$ .

[L1]

Continuity of $f$ at $x \in A$ : for every real $\varepsilon > 0$ there is a real $\delta > 0$ with $|f(y) - f(x)| < \varepsilon$ for every $y \in A$ satisfying $|y - x| < \delta$ ; equivalently $f\bigl(A \cap N_{\delta}(x)\bigr) \subseteq N_{\varepsilon}(f(x))$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point, The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L2]

Open sets: $V \subseteq \mathbb{R}$ is open when every point of $V$ has a neighbourhood inside $V$ , and every neighbourhood $N_{\delta}(x)$ is itself an open set containing $x$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L3]

Compactness of $K$ : for every family $\mathcal{W}$ of open subsets of $\mathbb{R}$ with $K \subseteq \bigcup \mathcal{W}$ , either $K = \varnothing$ and the empty subfamily covers it, or there are $n \in \mathbb{N}$ and $W_0, \dots, W_n \in \mathcal{W}$ with $K \subseteq W_0 \cup \dots \cup W_n$ (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset).

Proof

technique · direct

1.1

If $K = \varnothing$ , then $f[K] = \varnothing$ , which is compact by [L3]. Hence suppose $K \ne \varnothing$ for the rest of the proof.

L3cases

1.2

Let $\mathcal{V}$ be a family of open subsets of $\mathbb{R}$ with $f[K] \subseteq \bigcup \mathcal{V}$ . Define $\mathcal{W} \;:=\; \bigl\{\, N_{\delta}(x) \ : \ x \in K,\ \delta \in \mathbb{R},\ \delta > 0,\ \text{and } f\bigl(K \cap N_{\delta}(x)\bigr) \subseteq V \text{ for some } V \in \mathcal{V} \,\bigr\}.$ This family is cut out by a property of the pair $(x,\delta)$ , so it is a set and nothing is selected in forming it; every member is an open subset of $\mathbb{R}$ by [L2].

2.1

$\mathcal{W}$ covers $K$ . Let $x \in K$ . Then $x \in A$ and $f(x) \in f[K] \subseteq \bigcup \mathcal{V}$ , so $f(x) \in V$ for some $V \in \mathcal{V}$ . As $V$ is open, [L2] gives a real $\varepsilon > 0$ with $N_{\varepsilon}(f(x)) \subseteq V$ , and [L1] gives a real $\delta > 0$ with $f\bigl(A \cap N_{\delta}(x)\bigr) \subseteq N_{\varepsilon}(f(x)) \subseteq V$ . Since $K \subseteq A$ we get $f\bigl(K \cap N_{\delta}(x)\bigr) \subseteq V$ , so $N_{\delta}(x) \in \mathcal{W}$ , and $x \in N_{\delta}(x)$ by [L2]. Hence $K \subseteq \bigcup \mathcal{W}$ .

step 1.2L1L2

3.1

By [L3] there are $n \in \mathbb{N}$ and members $W_0, \dots, W_n$ of $\mathcal{W}$ with $K \subseteq W_0 \cup \dots \cup W_n$ . For each $i \le n$ the membership $W_i \in \mathcal{W}$ asserts the existence of some $V \in \mathcal{V}$ with $f(K \cap W_i) \subseteq V$ ; naming one such $V_i$ for each of the finitely many indices $i \le n$ requires no choice principle.

step 1.1step 1.2step 2.1L3choose

4.1

The finite subfamily works. Let $z \in f[K]$ , say $z = f(x)$ with $x \in K$ . By step 3.1 there is $i \le n$ with $x \in W_i$ , hence $x \in K \cap W_i$ and $z = f(x) \in V_i$ . Therefore $f[K] \subseteq V_0 \cup \dots \cup V_n$ , a union of finitely many members of $\mathcal{V}$ .

step 3.1

5.1

Every family of open subsets of $\mathbb{R}$ covering $f[K]$ thus has a finite subfamily covering $f[K]$ , so $f[K]$ is compact.

step 1.1step 1.2step 4.1L3∎

Remarks

The proof is the preimage characterisation, made canonical. $f : A \to \mathbb{R}$ is continuous on $A$ if and only if the preimage of every open subset of $\mathbb{R}$ is the intersection with $A$ of an open subset of $\mathbb{R}$ , and dually for closed sets says that $f^{-1}(V)$ is the trace on $A$ of an open set; the textbook argument pulls back each $V \in \mathcal{V}$ to such an open set and covers $K$ with those. Doing it that way selects one open set per member of $\mathcal{V}$ , a family that may be arbitrarily large. Uniting all admissible balls instead, as above, reaches the same cover with no selection, and the only naming step left is over a finite index set.
The image of a compact set is closed and bounded, by A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded. That is what the next two items use, and it is the shortest route to both boundedness and the attainment of extrema.
Compactness cannot be weakened to closedness or to boundedness. The image of a closed set need not be closed (FALSE: the image of a closed subset of $\mathbb{R}$ under a continuous real function is closed), and the image of a bounded set need not be bounded, since $x \mapsto 1/x$ carries the bounded set $(0,1)$ onto $(1,\infty)$ .

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 32 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Compact space (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 4 (Thm 4.14) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.3 (standard reference, not scraped)
W. Trench, Introduction to Real Analysis, Ch. 8: Metric Spaces (standard reference, not scraped)
MIT 18.100B lecture notes (standard reference, not scraped)

The image of a compact subset of R\mathbb{R}R under a continuous real function is compact