False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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FALSE: the image of a closed subset of $\mathbb{R}$ under a continuous real function is closed

Statement

False claim: if $A \subseteq \mathbb{R}$ , if $f : A \to \mathbb{R}$ is 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 if $F \subseteq A$ is a closed subset of $\mathbb{R}$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen), then the image $f[F] = \{\, f(x) : x \in F \,\}$ is a closed subset of $\mathbb{R}$ .

Why it is tempting. Continuity is characterised by the behaviour of preimages: the preimage of every closed set is relatively closed ( $f : A \to \mathbb{R}$ 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). It is easy to transpose that to images, and images are exactly where the characterisation says nothing.

What is true. Compactness, not closedness, is preserved: the image of a compact set under a continuous function is compact (The image of a compact subset of $\mathbb{R}$ under a continuous real function is compact), hence closed and bounded (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded). Closedness by itself is preserved by neither images nor unions of infinitely many closed sets, and boundedness by itself is not preserved either, since $x \mapsto 1/x$ carries the bounded set $(0,1)$ onto the unbounded set $(1,\infty)$ .

Facts & Assumptions

Given: The domain $A := \mathbb{R}$ , the closed set $F := \mathbb{R}$ , and the function $f : \mathbb{R} \to \mathbb{R}$ , $f(x) := 1/(1+x^{2})$ (Integer powers $a^m$ ).

[L1]

$\mathbb{R}$ is a closed subset of $\mathbb{R}$ , since its complement $\varnothing$ is open (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L2]

Algebra of continuous functions: polynomial functions are continuous on $\mathbb{R}$ , and if $q$ is continuous and nowhere zero on a set then $1/q$ is continuous there (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, 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).

[L3]

Squares and order: $x^{2} \ge 0$ for every real $x$ , so $1 + x^{2} \ge 1 > 0$ ; and $0 < s \le t$ implies $0 < 1/t \le 1/s$ (Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$ , Inverses of positives are positive, and reciprocation reverses order, Ordered field, Integer powers $a^m$ ).

[L4]

Square roots: every real $t \ge 0$ has a unique $s \ge 0$ with $s^{2} = t$ (Existence and uniqueness of $n$ -th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ ).

[L5]

Closure: $x \in \overline{S}$ exactly when $N_{\varepsilon}(x) \cap S \ne \varnothing$ for every real $\varepsilon > 0$ ; and $S$ is closed exactly when $S = \overline{S}$ (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ , Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ , The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L6]

Intervals and minima: $(0,1] = \{\, y : 0 < y \le 1 \,\}$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length); the minimum of a two-element set of reals exists and is one of them (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set); and $|u| \ge 0$ with $|u| = u$ for $u \ge 0$ (Basic properties of the absolute value).

Refutation

technique · direct

1.1

$F = \mathbb{R}$ is a closed subset of $\mathbb{R}$ and is contained in $A = \mathbb{R}$ .

1.2

$f$ is continuous on $\mathbb{R}$ : the denominator $x \mapsto 1 + x^{2}$ is a polynomial function, hence continuous by [L2], and it satisfies $1 + x^{2} \ge 1 > 0$ by [L3], so it never vanishes and $f = 1/(1+x^{2})$ is continuous by [L2].

L2L3

1.3

$(0,1] \subseteq f[\mathbb{R}]$ . Let $y$ satisfy $0 < y \le 1$ and put $t := 1/y - 1$ . By [L3] we have $1/y \ge 1$ , so $t \ge 0$ , and [L4] supplies a real $s \ge 0$ with $s^{2} = t$ . Then $1 + s^{2} = 1/y$ and hence $f(s) = 1/(1+s^{2}) = y$ .

L3L4L6

1.4

$(0,1]$ is not closed. Let a real $\varepsilon > 0$ be given and put $y := \min\{\varepsilon/2,\ 1\}$ , a real with $0 < y \le 1$ by [L6], so $y \in (0,1]$ ; and $|y - 0| = y \le \varepsilon/2 < \varepsilon$ , so $y \in N_{\varepsilon}(0) \cap (0,1]$ . Hence $0 \in \overline{(0,1]}$ by [L5], while $0 \notin (0,1]$ because $0 < 0$ is false. So $(0,1] \ne \overline{(0,1]}$ and $(0,1]$ is not closed by [L5].

L5L6

2.1

$f[\mathbb{R}] \subseteq (0,1]$ . For every real $x$ , [L3] gives $1 + x^{2} \ge 1 > 0$ and hence $0 < 1/(1+x^{2}) \le 1$ , that is $f(x) \in (0,1]$ .

step 1.2L3L6

3.1

So $f[F] = f[\mathbb{R}] = (0,1]$ .

step 2.1step 1.3

4.1

The set $F = \mathbb{R}$ is closed, $f$ is continuous on $\mathbb{R}$ , and $f[F] = (0,1]$ is not closed: the claim is false.

step 1.1step 1.2step 3.1step 1.4∎

Remarks

The witness is as tame as possible. $f$ is a quotient of polynomials, defined on the whole line, bounded, and its image is an interval; the failure is only that the infimum $0$ of the image is approached and not attained, because the points that would attain it have escaped to infinity. Replacing $\mathbb{R}$ by any closed unbounded set on which $f$ has infimum $0$ , such as $[0,\infty)$ , gives the same conclusion.
The image of a closed bounded set is closed, because such a set is compact (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded) and compactness is preserved (The image of a compact subset of $\mathbb{R}$ under a continuous real function is compact). So the false claim becomes true exactly when the hypothesis is strengthened from closed to compact, which is what The image of an interval under a continuous real function is order-convex, hence an interval, and the image of a closed bounded interval is a closed bounded interval uses in its second half.
Openness is not preserved either, in the other direction: the image of the open set $\mathbb{R}$ under this same $f$ is $(0,1]$ , which is not open, and the image of $\mathbb{R}$ under a constant function is a single point. Continuity constrains preimages, not images; that asymmetry is the content of $f : A \to \mathbb{R}$ 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.

Depends on

Used by

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Sources

Closed set (Wikipedia) (standard reference, not scraped)
Continuous function (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 4 (standard reference, not scraped)
MIT 18.100B lecture notes (standard reference, not scraped)

FALSE: the image of a closed subset of R\mathbb{R}R under a continuous real function is closed