TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$f$ is upper semicontinuous on $A$ if and only if $\{x \in A : f(x) < \alpha\}$ is relatively open in $A$ for every real $\alpha$ , lower semicontinuous if and only if $\{x \in A : f(x) > \alpha\}$ is, and continuous if and only if it is both

Statement

Let $A \subseteq \mathbb{R}$ and let $f : A \to \mathbb{R}$ . Call $S \subseteq A$ relatively open in $A$ when $S = U \cap A$ for some open $U \subseteq \mathbb{R}$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen). Then:

$f$ is upper semicontinuous on $A$ (Upper and lower semicontinuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ ) if and only if $\{\, x \in A : f(x) < \alpha \,\}$ is relatively open in $A$ for every real $\alpha$ ;
$f$ is lower semicontinuous on $A$ if and only if $\{\, x \in A : f(x) > \alpha \,\}$ is relatively open in $A$ for every real $\alpha$ ;
$f$ 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) if and only if both families of sets are relatively open.

The open set is produced canonically, not chosen. For each $\alpha$ the proof exhibits one specific open $U_\alpha \subseteq \mathbb{R}$ with $U_\alpha \cap A = \{f < \alpha\}$ , namely the set of reals $y$ admitting a radius $\rho$ with $A \cap N_\rho(y) \subseteq \{f < \alpha\}$ . No choice of a radius per point is made, which matters because the level set may be uncountable.

Facts & Assumptions

Given: $A \subseteq \mathbb{R}$ and a function $f : A \to \mathbb{R}$ .

[L1]

$f$ is upper semicontinuous at $c \in A$ when for every real $\varepsilon > 0$ there is a real $\delta > 0$ with $f(x) < f(c) + \varepsilon$ for every $x \in A \cap N_\delta(c)$ ; lower semicontinuity is the same with $f(x) > f(c) - \varepsilon$ ; and continuity at $c$ is the conjunction of the two (Upper and lower semicontinuity of $f : A \to \mathbb{R}$ at a point of $A$ and 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).

[L2]

$U \subseteq \mathbb{R}$ is open exactly when every $y \in U$ has a real $\rho > 0$ with $N_\rho(y) \subseteq U$ ; and if $|z - y| < \rho/2$ then $N_{\rho/2}(z) \subseteq N_\rho(y)$ (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]

$-f$ is lower semicontinuous at $c$ exactly when $f$ is upper semicontinuous at $c$ , and $\{x \in A : -f(x) < \alpha\} = \{x \in A : f(x) > -\alpha\}$ (Upper and lower semicontinuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ ).

Proof

technique · direct

1.1

Fix a real $\alpha$ and put $S_\alpha := \{\, x \in A : f(x) < \alpha \,\}$ and $U_\alpha := \{\, y \in \mathbb{R} : A \cap N_\rho(y) \subseteq S_\alpha \ \text{for some real } \rho > 0 \,\}$ .

construct

1.2

Conversely suppose every $S_\alpha$ is relatively open in $A$ , say $S_\alpha = U \cap A$ with $U$ open, and let $c \in A$ and $\varepsilon > 0$ be real. Put $\alpha := f(c) + \varepsilon$ ; then $f(c) < \alpha$ , so $c \in S_\alpha = U \cap A$ , and there is a real $\delta > 0$ with $N_\delta(c) \subseteq U$ .

2.1

$U_\alpha$ is open: if $y \in U_\alpha$ with witness $\rho$ and $z \in N_{\rho/2}(y)$ , then $A \cap N_{\rho/2}(z) \subseteq A \cap N_\rho(y) \subseteq S_\alpha$ , so $z \in U_\alpha$ with witness $\rho/2$ ; hence $N_{\rho/2}(y) \subseteq U_\alpha$ .

step 1.1L2

2.2

$U_\alpha \cap A \subseteq S_\alpha$ : if $y \in U_\alpha \cap A$ with witness $\rho$ then $y \in A \cap N_\rho(y) \subseteq S_\alpha$ .

step 1.1

2.3

Suppose $f$ is upper semicontinuous on $A$ and let $c \in S_\alpha$ . Apply the definition at $c$ with $\varepsilon := \alpha - f(c) > 0$ : there is a real $\delta > 0$ with $f(x) < f(c) + \varepsilon = \alpha$ for every $x \in A \cap N_\delta(c)$ , that is $A \cap N_\delta(c) \subseteq S_\alpha$ ; so $c \in U_\alpha$ .

step 1.1L1

3.1

Hence $S_\alpha \subseteq U_\alpha \cap A$ , and with step 2.2 this gives $S_\alpha = U_\alpha \cap A$ , a relatively open subset of $A$ ; since $\alpha$ was arbitrary, one direction of claim 1 holds.

step 2.2step 2.3

4.1

With $\delta$ as in step 1.2, every $x \in A \cap N_\delta(c)$ lies in $U \cap A = S_\alpha$ , so $f(x) < \alpha = f(c) + \varepsilon$ . As $c$ and $\varepsilon$ were arbitrary, $f$ is upper semicontinuous on $A$ , which completes claim 1.

step 3.1step 1.2L1

5.1

Claim 2 follows by applying claim 1 to $-f$ : $f$ is lower semicontinuous on $A$ exactly when $-f$ is upper semicontinuous on $A$ , exactly when $\{x \in A : -f(x) < \beta\}$ is relatively open for every real $\beta$ , and that set is $\{x \in A : f(x) > -\beta\}$ ; as $\beta$ ranges over the reals so does $-\beta$ .

step 4.1L3

6.1

Claim 3 follows: $f$ is continuous on $A$ exactly when it is both upper and lower semicontinuous on $A$ , and by claims 1 and 2 that is exactly the conjunction of the two families of sets being relatively open.

step 4.1step 5.1L1∎

Remarks

Why "relatively" open and not open. $S_\alpha$ is a subset of $A$ , so it cannot be open in $\mathbb{R}$ unless $A$ is; the correct statement is the one above, exactly as in $f : A \to \mathbb{R}$ 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, where the same phrase is fixed inline for the same reason. For $A = \mathbb{R}$ the qualifier disappears and the level sets are open outright.
The strict inequalities are not interchangeable with the weak ones. Upper semicontinuity says the strict sublevel sets are relatively open, equivalently that the sets $\{f \ge \alpha\}$ are relatively closed. It does not say that the sets $\{f \le \alpha\}$ are relatively closed; the indicator of a closed set is upper semicontinuous while $\{f \le 0\}$ is the complement of that closed set, which is relatively open and generally not closed.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 26 results over 12 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

Semi-continuity (Wikipedia) (standard reference, not scraped)
Characterization of Lower Semicontinuity (ProofWiki) (standard reference, not scraped)

fff is upper semicontinuous on AAA if and only if {x∈A:f(x)<α}\{x \in A : f(x) < \alpha\}{x∈A:f(x)<α} is relatively open in AAA for every real α\alphaα, lower semicontinuous if and only if {x∈A:f(x)>α}\{x \in A : f(x) > \alpha\}{x∈A:f(x)>α} is, and continuous if and only if it is both