Continuity of the cyclic block-minimum value

Fix a parameter $a$, let $u=T_i(a)$, and put $m_i(a)=f(a,u)+g_i(u)$. For $\epsilon >0$, set $c=g_i(u)-\epsilon \mathrm{/}3$. The weak sublevel $C_c=\{v:g_i(v)\le c\}$ is closed because $v\mapsto (v,c)$ is a continuous slice of the closed epigraph in [F2]; since $u\mathrm{\notin }{C}_c$, its open complement contains a neighborhood of $u$ on which $g_i>c$. Continuity of $T_i$ and $f$ gives a relative neighborhood $V$ of $a$ on which $T_i(a^{\mathrm{\prime }})$ lies in that neighborhood and both $f(a^{\mathrm{\prime }},T_i(a^{\mathrm{\prime }}))$ and $f(a^{\mathrm{\prime }},u)$ differ from $f(a,u)$ by less than $\epsilon \mathrm{/}3$. Hence $m_i(a^{\mathrm{\prime }})>m_i(a)-2\epsilon \mathrm{/}3$.

For each $a^{\mathrm{\prime }}\in V$, the point $u$ is a competitor, so $m_i(a^{\mathrm{\prime }})\le f(a^{\mathrm{\prime }},u)+g_i(u)<m_i(a)+\epsilon \mathrm{/}3$. Together with step 1.1, this gives $\mathrm{\mid }{m}_i(a^{\mathrm{\prime }})-m_i(a)\mathrm{\mid }<\epsilon$, which is continuity of $m_i$ at the arbitrary parameter $a$. The argument uses neighborhoods and closed epigraphs directly; it does not infer topological continuity from a sequential criterion.