Uniform limit of real valued functions on a compact space. Is the union of their images necessarily compact? Let $K$ be a compact space with $f_n$, $f$ continuous functions $K \to \mathbb{R}$ such that $f_n \to f$ uniformly.
Is $\mathrm{im}f \cup \bigcup \mathrm{im}(f_n) \subseteq \mathbb{R}$ necessarily compact?
For $\varepsilon = 1$ we get an $N$ such that $|f_n - f|$ < 1 for all $n > N$.
$\mathrm{im}(f) \subseteq [a,b]$ and $\mathrm{im}(f_n) \subseteq [a_n,b_n]$ for some $a,b, a_n, b_n \in \mathbb{R}$.
Letting $a^*=\mathrm{min}(a-1,a_1,...,a_N)$ and $b^*=\mathrm{min}, b^* = (b+1,b_1,...,b_N)$, we see that $\mathrm{im}f \cup \bigcup \mathrm{im}(f_n) \subseteq [a^*,b^*]$.
Must the union of images be closed?

If I am not mistaken, another kind of solution.
a) Let $E$ be the space of continuous fonctions on $K$ with values in $\mathbb{R}$, equipped with the "Sup" norm. Let $F=E\times K$. Then the application $\phi$: $F\to \mathbb{R}$ defined by $\phi((g,y))=g(y)$ is continuous. To see why, note that if $(g_0, y_0)$ is given, and $\varepsilon>0$ 
we have
$$|\phi((g,y))-\phi((g_0,y_0))|\leq \|g-g_0\|+|g_0(y)-g_0(y_0)|$$
As there exist a neigbourhood $V$ of $y_0$ such that for $y\in V$, we have $|g_0(y)-g_0(y_0)|<\varepsilon/2$ (continuity of $g_0$), we get for $(g,y)\in B(g_0,\varepsilon/2)\times V$ that $\displaystyle |\phi((g,y))-\phi((g_0,y_0))|<\varepsilon$.
b) As 
$f_n\to f$ in $E$, the set $A=\{f_n, n\geq 1,f\}$ is compact in $E$.
c) Hence $A\times K=B$ is compact in $F$, and $\phi(B)$ is compact as a continuous image of a compact, and we are done. 
