[proofplan] We set $m=\inf_{z\in\mathcal A}J[z]$ and distinguish the degenerate case $m=\infty$ from the finite case. If $m<\infty$, the definition of the infimum produces a minimizing sequence in $\mathcal A$. [Sequential compactness](/page/Sequential%20Compactness) gives a $\tau$-convergent subsequence with limit $z^*\in\mathcal A$, and sequential lower semicontinuity gives $J[z^*]\le m$. Since $m$ is a lower bound for all values of $J$, the reverse inequality follows, so $z^*$ attains the infimum. [/proofplan] [step:Reduce the problem to the finite infimum case] Define the extended real number $m\in\mathbb R\cup\{\infty\}$ by \begin{align*}m:=\inf_{z\in\mathcal A}J[z].\end{align*} Because $J$ is bounded below, there exists a real number $c\in\mathbb R$ such that $c\le J[z]$ for every $z\in\mathcal A$. Hence $m\ne -\infty$. If $m=\infty$, then $J[z]=\infty$ for every $z\in\mathcal A$. Since $\mathcal A$ is nonempty, choose $z^*\in\mathcal A$. Then \begin{align*}J[z^*]=\infty=m=\inf_{z\in\mathcal A}J[z].\end{align*} so $z^*$ is a minimizer. It remains to consider the case $m\in\mathbb R$. [/step] [step:Construct a minimizing sequence from the definition of the infimum] Assume $m\in\mathbb R$. For each $k\in\mathbb N$, the number $m+1/k$ is strictly larger than $m$. Since $m$ is the greatest lower bound of the set $\{J[z]:z\in\mathcal A\}\subset\mathbb R\cup\{\infty\}$, $m+1/k$ cannot be a lower bound for this set. Therefore there exists $z_k\in\mathcal A$ such that \begin{align*} J[z_k]<m+\frac{1}{k}. \end{align*} Also, since $m$ is a lower bound, $m\le J[z_k]$ for every $k\in\mathbb N$. Thus \begin{align*} m\le J[z_k]<m+\frac{1}{k}. \end{align*} By the squeeze property for real sequences, $J[z_k]\to m$ as $k\to\infty$. [guided] We need a sequence in $\mathcal A$ whose $J$-values approach the best possible value $m$. Since $m$ is an infimum, it may not be attained a priori; the point of the argument is to prove that compactness and lower semicontinuity force attainment. Fix $k\in\mathbb N$. The real number $m+1/k$ is larger than $m$. If $m+1/k$ were still a lower bound for all values $J[z]$, then $m$ would not be the greatest lower bound. Hence there must be at least one element $z_k\in\mathcal A$ with \begin{align*} J[z_k]<m+\frac{1}{k}. \end{align*} On the other hand, $m$ itself is a lower bound for the values of $J$, so the same point satisfies \begin{align*} m\le J[z_k]. \end{align*} Combining the two inequalities gives \begin{align*} m\le J[z_k]<m+\frac{1}{k}. \end{align*} The error term $1/k$ tends to $0$, so the squeeze property gives \begin{align*} \lim_{k\to\infty}J[z_k]=m. \end{align*} Thus $(z_k)_{k\in\mathbb N}$ is a minimizing sequence in $\mathcal A$. [/guided] [/step] [step:Extract a convergent subsequence inside the admissible set] The sequence $(z_k)_{k\in\mathbb N}$ lies in $\mathcal A$. Since $(\mathcal A,\tau)$ is sequentially compact, there exist a strictly increasing map $j\mapsto k_j$ from $\mathbb N$ to $\mathbb N$ and a point $z^*\in\mathcal A$ such that \begin{align*}z_{k_j}\to z^* \quad \text{with respect to }\tau.\end{align*} Because every subsequence of a convergent real sequence has the same limit, the previously established convergence $J[z_k]\to m$ implies \begin{align*}\lim_{j\to\infty}J[z_{k_j}]=m.\end{align*} [/step] [step:Use sequential lower semicontinuity to compare the limit value with the infimum] Sequential lower semicontinuity of $J$ with respect to $\tau$ applies to the sequence $(z_{k_j})_{j\in\mathbb N}$ because $z_{k_j}\to z^*$ in $\tau$. Therefore \begin{align*}J[z^*]\le \liminf_{j\to\infty}J[z_{k_j}].\end{align*} Since $J[z_{k_j}]\to m$, its limit inferior equals $m$, and hence \begin{align*}J[z^*]\le m.\end{align*} [/step] [step:Conclude that the subsequential limit attains the minimum] By the definition of $m$ as the infimum of $J$ over $\mathcal A$, $m$ is a lower bound for all values $J[z]$ with $z\in\mathcal A$. Since $z^*\in\mathcal A$, this gives \begin{align*}m\le J[z^*].\end{align*} Together with $J[z^*]\le m$, we obtain \begin{align*}J[z^*]=m=\inf_{z\in\mathcal A}J[z].\end{align*} Thus $z^*$ is a minimizer of $J$ over $\mathcal A$. [/step]