$f(K) \cup \left(\bigcup_{i=1}^\infty f_i(K)\right)$ is compact Let $f_i:\mathbb{R}^n\to \mathbb{R}^m$ be continuous for all $i\in \mathbb{N},$ and $K$ be any compact subset of $\mathbb{R}^n$. Suppose that $f_i\to f$ uniformly on $K$. Prove that the set $$ f(K) \cup \left(\bigcup_{i=1}^\infty f_i(K)\right) $$ is compact. 
What I know is $f(K)$ is compact and for $I\in \mathbb{N}$ $f_i(K)$ is also compact. Now how will I conclude that given set is compact.
 A: Your set is bounded; this follows from the fact that $f(K)$ is bounded and from the uniform convergence of $(f_n)_{n\in\mathbb N}$. So, all that is need is to prove that it is closed.
If your set was not closed, there would be a sequence of points of it that would converge to a point outside of it. Let $f_0=f$ and take a sequence $(x_n)_{n\in\mathbb N}$ of points of $K$ and a sequence $(i_n)_{n\in\mathbb N}$ of non-negative integers. Suppose that $\lim_{n\to\infty}f_{i_n}(x_n)=y$. I will prove that $y\in\bigcup_{n\in\mathbb{Z}^+}f_n(K)$.
Since $K$ is compact, you can assume, without loss of generality, that the sequence $(x_n)_{n\in\mathbb N}$ converges to some $x\in K$. If every $i_n$ is equal to $0$ after a certain point, then $y=f_0(x)=f(x)$. So, again  without loss of generality, we can assume that $(\forall n\in\mathbb{N}):i_n\in\mathbb N$. And if, for some $k\in\mathbb N$, $i_n=k$ infinitely many times, then $y=f_k(x)$. So, we can assume that, for each natural $k$, the equality $i_n=k$ occurs only finitely many times. But this implies that, if $n$ is large enough, then $i_n$ will also be large.
Now, it's time to use uniform continuity. Take $\varepsilon>0$. If $n$ is large enough, then$$(\forall k\in K):\bigl|f(k)-f_n(k)\bigr|<\varepsilon.$$Therefore, if $n$ is large enough, then $\bigl|f(x_n)-f_{i_n}(x_n)\bigr|<\varepsilon$. But then$$y=\lim_{n\to\infty}f_{i_n}(x_n)=\lim_{n\to\infty}f(x_n)\in f(K).$$
A: You could consider the set $\mathbb{N}_{\infty} = \mathbb{N} \cup \{\infty\}$ with the topology $\tau$ as follows: a subset $U$ of $\mathbb{N}_{\infty}$ is open iff $U \subset \mathbb{N}$ or $U \ni \infty$ and $\mathbb{N} \backslash U$ is finite. This is really the one-point compactification of ($\mathbb{N}$ with the discrete topology). Now, let's define $F\colon \mathbb{N}_{\infty} \times \mathbb{R}^n \to \mathbb{R}^m$, $F(n,x) = f_n(x)$ and $F(\infty, x) = f(x)$. Then $f_n$ converges uniformly to $f$ implies $F$ is continuous. Now, one only has to notice that the union in the OP is $F(\mathbb{N}_{\infty} \times K)$, an image of a compact set. 
If one does not want to go through all these fancy moves, at least we know how the proof might work. Start with a sequence $f_{n_k}(x_k)$ and show that it has a convergent subsequence (we see now a reason for the index $\infty$). Anyways, $n_k$ will have a convergent subsequence ( either constant, or to $\infty$), then out of that, take another subsequence so the $x$ component converges. The reasoning is now standard. 
A: Hint: give a look at the distance between $f_i(K)$ and $f(K)$.
