Does uniform convergence on the complement of a one point set imply uniform convergence? Two different function sequences $\{f_n\}$ and $\{g_n\}$ both converge to a function $h : \mathbb{R} \to \mathbb{R}$ uniformly everywhere in $(a,b)$ except at $x = c \in (a,b)$ where they converge non uniformly. I'd like to know if $\{f_n - g_n\}$ converges to zero uniformly everywhere ?
 A: I'm not sure how you can have uniform convergence except at a single point.  Let $c$ be the point, and let $U=(a,b)\setminus \{c\}$.  Because we converge uniformly on $U$ and pointwise at $c$, we have that for all $\epsilon>0$, we have $N_{\epsilon}$ and $M_{\epsilon}$ such that if $n>N_{\epsilon}$, then $|f(y)-f_n(y)|<\epsilon$ for all $y\in U$ and if $n>M_{\epsilon}$, then $|f(c)-f_n(c)|<\epsilon$.  However, if we use $N=\max(M_{\epsilon},N_{\epsilon})$, then $|f_n(y)-f(y)|<\epsilon$ for EVERY $y\in (a,b)$ (including $c$) when $n>N$, and therefore we converge uniformly.
Or is that not what you meant by uniform convergence except at $c$?

More generally, we can extend this argument to show that if $U= \displaystyle\bigcup_{1\leq i \leq n} U_i$ and $f$ converges uniformly on each $U_i$, then $f$ converges uniformly on $U$.  If $U_i$ is a single point, then pointwise convergence on $U_i$ is the same as uniform convergence.
A: The short answer is yes.
Actually if $f_n \rightarrow h$ on $(a,b)$ and uniformly on $(a,c)\cup(c,b)$, then $f_n \rightarrow h$ uniformly on $(a,b)$.
To see this, we invoke the definition of uniform convergence $f_n \rightarrow h$:
$\forall \epsilon>0: \exists N_\epsilon: n\geq N_\epsilon \Rightarrow |f_n-h|_\infty<\epsilon$
If this holds on $(a,c)\cup(c,b)$, and convergence holds on $c$:
$\forall \epsilon>0: \exists M_\epsilon: m\geq M_\epsilon \Rightarrow |f_m(c)-h(c)|<\epsilon$
Then picking $K_\epsilon = \max(N_\epsilon,M_\epsilon)$ establishes:
$\forall \epsilon>0: \exists K_\epsilon: k\geq K_\epsilon \Rightarrow |f_k-h|_\infty<\epsilon$
on $(a,b)$ and therefore proves uniform convergence on $(a,b)$.
So we have $f_n$ and $g_n$ both converging uniformly to $h$ on $(a,b)$. Thus:
$\forall \epsilon>0: \exists K_\epsilon: k\geq K_\epsilon \Rightarrow |f_k-h|_\infty<\epsilon$
and
$\forall \epsilon>0: \exists L_\epsilon: l\geq L_\epsilon \Rightarrow |g_l-h|_\infty<\epsilon$
Now picking $P_\epsilon = \max(K_\frac{\epsilon}{2},L_\frac{\epsilon}{2})$ gives:
$\forall \epsilon>0: \exists P_\epsilon: p\geq P_\epsilon \Rightarrow |f_p-h|_\infty<\frac{\epsilon}{2}$ and $|g_p-h|_\infty<\frac{\epsilon}{2}$ and thus by the triangle inequality $|(f_p-h)+(h-g_p)|_\infty = |(f_p-g_p)-0|_\infty < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$
So we've proved uniform convergence of $f_p-g_p$ to $0$ on $(a,b)$.
