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I can partially answer the second question. If $X$ is a compact Hausdorff space whose topology has a countable base at every point (e.g. [Edit: $\omega_1$), \omega_1$has this property but not compact], then there are no bad sequences. Moreover the following holds: If$F\subset C(X)$is a family such that every countable subfamily has radius$\le 1$, then$r(F)\le 1$. Define a function$S=S_F:X\to\widehat{\mathbb R}=[-\infty,+\infty]$(the "essential supremum" of$F$) as follows:$S(x)$is the maximum$t$such that for every neighborhood$U$of$x$one has $\sup\{f(y):f\in F,y\in U\}\ge t$. Observe that$S$is upper semi-continuous: for every$t\in\widehat{\mathbb R}$, the set $\{x\in X:S(x)\ge t\}$ is closed. Define the essential infimum$I_F$similarly, this function is lower semi-continuous. For every$x\in X$there is a countable family$G\subset F$such that$S_G(x)=S_F(x)$and$I_G(x)=I_F(x)$. Indeed, using countable base at$x$, one can realize$S(x)$by a sequence$x_i:i\in\mathbb N$converging to$x$and functions$f_i\in F$such that$f_i(x_i)\to S(x)$. It follows that$S(x)\le I(x)+2$for all$x\in X$. Indeed, take$G$as above, it is contained in a$(1+\epsilon)$-ball centered at some$f\in C(X)$, then$S_G(x)\le f(x)+1+\epsilon$and$I_G(x)\ge f(x)-1-\epsilon$. Fix$\epsilon>0$and let us prove that$F$is contained in a$(1+\epsilon)$-ball. For$x\in X$, define$C_x=\frac12(S(x)+I(x))$. Note that$S(x)\le C_x+1$and$I(x)\ge C_x-1$. By semi-continuity, there is a neighborhood$U_x$of$x$such that $S(y)<C_x+1+\epsilon$ and $I(y)>C_x-1-\epsilon$ for all$y\in U_x$. Choose a finite subcovering $V_i=U_{x_i}:i\le N$. On each neighborhood$V_i$we have a constant function$f_i:=C_{x_i}$which works as a center within this neighborhood. It suffices to construct a function$g\in C(X)$such that for every$x\in X$,$g(x)$is between the minimum and maximum of these partially defined constant functions at$x$. This is easy to do by induction in the number of sets in the covering. Suppose we have already defined$g=g_{n-1}$that works on $\bigcup_{i<n} V_i$. By Urysohn's lemma there is$\phi:X\to[0,1]$such that$\phi=0$on$X\setminus V_n$and$\phi=1$on$X\setminus\bigcup_{i\ne n} V_i$. Then$g_n:=\phi f_n+(1-\phi)g_{n-1}$works on$\bigcup_{i\le n}V_i$. 1 I can partially answer the second question. If$X$is a compact Hausdorff space whose topology has a countable base at every point (e.g.$\omega_1$), then there are no bad sequences. Moreover the following holds: If$F\subset C(X)$is a family such that every countable subfamily has radius$\le 1$, then$r(F)\le 1$. Define a function$S=S_F:X\to\widehat{\mathbb R}=[-\infty,+\infty]$(the "essential supremum" of$F$) as follows:$S(x)$is the maximum$t$such that for every neighborhood$U$of$x$one has $\sup\{f(y):f\in F,y\in U\}\ge t$. Observe that$S$is upper semi-continuous: for every$t\in\widehat{\mathbb R}$, the set $\{x\in X:S(x)\ge t\}$ is closed. Define the essential infimum$I_F$similarly, this function is lower semi-continuous. For every$x\in X$there is a countable family$G\subset F$such that$S_G(x)=S_F(x)$and$I_G(x)=I_F(x)$. Indeed, using countable base at$x$, one can realize$S(x)$by a sequence$x_i:i\in\mathbb N$converging to$x$and functions$f_i\in F$such that$f_i(x_i)\to S(x)$. It follows that$S(x)\le I(x)+2$for all$x\in X$. Indeed, take$G$as above, it is contained in a$(1+\epsilon)$-ball centered at some$f\in C(X)$, then$S_G(x)\le f(x)+1+\epsilon$and$I_G(x)\ge f(x)-1-\epsilon$. Fix$\epsilon>0$and let us prove that$F$is contained in a$(1+\epsilon)$-ball. For$x\in X$, define$C_x=\frac12(S(x)+I(x))$. Note that$S(x)\le C_x+1$and$I(x)\ge C_x-1$. By semi-continuity, there is a neighborhood$U_x$of$x$such that $S(y)<C_x+1+\epsilon$ and $I(y)>C_x-1-\epsilon$ for all$y\in U_x$. Choose a finite subcovering $V_i=U_{x_i}:i\le N$. On each neighborhood$V_i$we have a constant function$f_i:=C_{x_i}$which works as a center within this neighborhood. It suffices to construct a function$g\in C(X)$such that for every$x\in X$,$g(x)$is between the minimum and maximum of these partially defined constant functions at$x$. This is easy to do by induction in the number of sets in the covering. Suppose we have already defined$g=g_{n-1}$that works on $\bigcup_{i<n} V_i$. By Urysohn's lemma there is$\phi:X\to[0,1]$such that$\phi=0$on$X\setminus V_n$and$\phi=1$on$X\setminus\bigcup_{i\ne n} V_i$. Then$g_n:=\phi f_n+(1-\phi)g_{n-1}$works on$\bigcup_{i\le n}V_i\$.