Proof of Sequential Characterization of Lower Semicontinuity on a Subset of a Metric Space

Throughout we use the order arithmetic of Elementary Order Arithmetic in an Ordered Field. Its clauses 1 and 10 are stated for strict inequalities; the corresponding statements for $\le$, and the transitivity of $\le$, follow by treating the equality case (and, for multiplication, the case of a zero multiplier) separately, and we use them under this convention without further comment. We write $\iota:\mathbb{N}\to\mathbb{R}$ for the canonical map of $\mathbb{R}$.

Claim 1, necessity. Assume $u$ is lower semicontinuous at $x$ relative to $A$. Let $(y_{j})_{j\in\mathbb{N}}$ be a sequence in $A$ converging to $x$ in $(A,d_{A})$ and let $\varepsilon$ be a real number with $0<\varepsilon$. By the definition of lower semicontinuity at $x$ there is a real $\delta>0$ such that every $y\in A$ with $d(x,y)<\delta$ satisfies $u(x)-\varepsilon<u(y)$. By the definition of convergence in $(A,d_{A})$ there is $N\in\mathbb{N}$ such that $d_{A}(y_{j},x)<\delta$ for every $j\ge N$. Since $d_{A}$ is the restriction of $d$ and a metric is symmetric, $d(x,y_{j})=d_{A}(x,y_{j})=d_{A}(y_{j},x)<\delta$ for every such $j$, and $y_{j}\in A$. Hence $u(x)-\varepsilon<u(y_{j})$ for every $j\ge N$, which is the stated condition.

Claim 1, sufficiency. We prove the contrapositive: assuming that $u$ is not lower semicontinuous at $x$ relative to $A$, we produce a sequence in $A$ converging to $x$ for which the condition fails.

By the negation of the definition there is a real $\varepsilon_{0}>0$ such that for every real $\delta>0$ there exists $y\in A$ with $d(x,y)<\delta$ for which $u(x)-\varepsilon_{0}<u(y)$ is false. Since $\le$ is a total order on $\mathbb{R}$, the failure of $u(x)-\varepsilon_{0}<u(y)$ means exactly that $u(y)\le u(x)-\varepsilon_{0}$.

For $j\in\mathbb{N}$, claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field gives $0<\iota(j)$, so $\iota(j)^{-1}$ exists and $0<\iota(j)^{-1}$. Put

$$A_{j}=\bigl\{y\in A\;:\;d(x,y)<\iota(j)^{-1}\ \text{ and }\ u(y)\le u(x)-\varepsilon_{0}\bigr\}.$$

Applying the previous paragraph with $\delta=\iota(j)^{-1}$ shows that each $A_{j}$ is nonempty, and every $A_{j}$ is a subset of the one set $A$. Hence Axiom of Countable Choice furnishes a sequence $(y_{j})_{j\in\mathbb{N}}$ with $y_{j}\in A_{j}$ for every $j\in\mathbb{N}$.

We check that $(y_{j})_{j\in\mathbb{N}}$ converges to $x$ in $(A,d_{A})$. Let $\varepsilon$ be a real number with $0<\varepsilon$. By claim 3 of The Archimedean Property of the Real Numbers there is $n\in\mathbb{N}$ with $0<\iota(n)^{-1}<\varepsilon$. Let $j\in\mathbb{N}$ with $j\ge n$. If $j=n$ then $\iota(j)^{-1}=\iota(n)^{-1}$. If $n<j$ then $\iota(n)<\iota(j)$ by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. In either case $\iota(n)\le\iota(j)$, and both numbers are positive. Multiplying $\iota(n)\le\iota(j)$ by the positive number $\iota(j)^{-1}$ gives $\iota(n)\,\iota(j)^{-1}\le\iota(j)\,\iota(j)^{-1}=1$, and multiplying that inequality by the positive number $\iota(n)^{-1}$ gives

$$\iota(j)^{-1}=\iota(n)^{-1}\,\iota(n)\,\iota(j)^{-1}\le\iota(n)^{-1}.$$

Since $y_{j}\in A_{j}$ we have $d_{A}(y_{j},x)=d(x,y_{j})<\iota(j)^{-1}\le\iota(n)^{-1}<\varepsilon$, using symmetry of the metric as above. Therefore $(y_{j})_{j\in\mathbb{N}}$ converges to $x$ in $(A,d_{A})$.

Finally, the condition of claim 1 fails for this sequence: taking $\varepsilon=\varepsilon_{0}$ in that condition would give $N\in\mathbb{N}$ with $u(x)-\varepsilon_{0}<u(y_{j})$ for every $j\ge N$; but $y_{N}\in A_{N}$ gives $u(y_{N})\le u(x)-\varepsilon_{0}$, whence $u(x)-\varepsilon_{0}<u(y_{N})\le u(x)-\varepsilon_{0}$, so $u(x)-\varepsilon_{0}<u(x)-\varepsilon_{0}$, which is impossible. This proves the contrapositive and completes claim 1.

Claim 2. Write $a=\liminf_{j}u(y_{j})$ and, for $k\in\mathbb{N}$, $A_{k}^{u}=\{u(y_{m}):m\in\mathbb{N},\ m\ge k\}$, so that by the definition of the limit inferior $a=\sup\{\inf A_{k}^{u}:k\in\mathbb{N}\}$.

Let $\varepsilon$ be a real number with $0<\varepsilon$. By claim 1 there is $N\in\mathbb{N}$ with $u(x)-\varepsilon<u(y_{j})$, hence $u(x)-\varepsilon\le u(y_{j})$, for every $j\ge N$. Thus $u(x)-\varepsilon$ is a lower bound of $A_{N}^{u}$, and since $\inf A_{N}^{u}$ is the greatest lower bound of $A_{N}^{u}$ we get $u(x)-\varepsilon\le\inf A_{N}^{u}$. As $\inf A_{N}^{u}$ belongs to the set $\{\inf A_{k}^{u}:k\in\mathbb{N}\}$ and $a$ is an upper bound of that set, $\inf A_{N}^{u}\le a$. Therefore

$$u(x)-\varepsilon\le a\qquad\text{for every real }\varepsilon>0. \tag{$*$}$$

Suppose, for contradiction, that $u(x)\le a$ fails. Since $\le$ is a total order this gives $a<u(x)$, so $0<u(x)-a$. By claim 8 of Elementary Order Arithmetic in an Ordered Field there is a real $\varepsilon_{1}$ with $0<\varepsilon_{1}$ and $\varepsilon_{1}+\varepsilon_{1}=u(x)-a$. Then $u(x)-\varepsilon_{1}=a+\varepsilon_{1}$, and $a<a+\varepsilon_{1}$ because $0<\varepsilon_{1}$. Applying $(*)$ with $\varepsilon=\varepsilon_{1}$ gives $a+\varepsilon_{1}\le a$, so $a<a+\varepsilon_{1}\le a$ and hence $a<a$, which is impossible. Therefore $u(x)\le a$, which is claim 2.

Claim 3. By the definition, $u$ is lower semicontinuous on $A$ precisely when it is lower semicontinuous at $x$ relative to $A$ for every $x\in A$. By claim 1 the latter holds at a given $x\in A$ precisely when the stated sequential condition holds at that $x$. Quantifying over $x\in A$ gives claim 3.