An exercise.

Exercise. Consider $\mathbb{R}^{\infty}=\bigcup_n \mathbb{R}^n$ with the colimit topology: $U\subseteq \mathbb{R}^{\infty}$ is open if and only if $U\cap \mathbb{R}^n$ is open for all $n$. Prove that if $C\subseteq \mathbb{R}^{\infty}$ is compact, then $C\subseteq\mathbb{R}^n$ for some $n\in\mathbb{N}$.

Proof. Let $\mathbb{R}^{\omega}$ be the product of countably many copies of $\mathbb{R}$. Note that we can identify $\mathbb{R}^{\infty}$ with the set $$A=\lbrace (x_1,x_2,\ldots)\in\mathbb{R}^{\omega} \,|\, \exists n\in \mathbb{N} \text{ such that }x_i=0\text{ for all }i\geq n\rbrace .$$

Consider the box topology on $\mathbb{R}^{\omega}$, $\tau_{box}$, and let $\tau_{sub}$ be the subspace topology on $\mathbb{R}^{\infty}$. Let us denote by $\tau_{col}$ the colimit topology on $\mathbb{R}^{\infty}$. It is clear that each element of the usual basis of $\tau_{sub}$ (inherited from $\tau_{box}$) is an element of $\tau_{col}$, thus $\tau_{sub}\subseteq \tau_{col}$.

Define a function $$i:\mathbb{R}^{\infty}\rightarrow\mathbb{N}$$ given by $$i(x)= \text{ Index of rightmost non-zero entry of }x.$$ We will now proceed by contradiction. Suppose that $C$ is not contained in any $\mathbb{R}^n$. With this assumption, we can find a sequence of elements $\lbrace x_n\rbrace_{n\geq 1} \subseteq C$ such that $i(x_n)<i(x_{n+1})$ $\forall$ $n\in \mathbb{N}$.

Put $i_l=i(x_l)$, $i_0=0$, and $p_l=|\pi_{i_l}(x_l)|$, where $\pi_{i}$ denotes the projection on the $i$-th coordinate, and define $$I=\prod_{l=1}^\infty\prod_{j={i_{l-1}}+1}^{i_l}(-\frac{p_l}{2},\frac{p_l}{2})\subset \mathbb{R}^{\omega},$$which belongs to $\tau_{box}$. Note that since $p_l$ is the rightmost (non-zero) entry of $x_l$ and it is in the $i_l$ coordinate, then $x_l\notin I$ $\forall$ $l$. Now, for each $n\in \mathbb{N}$, define the set $$\bar{U}_n=\prod_{j=1}^{n}\mathbb{R}\times \prod_{j=n+1} I_j,$$ where $I_j$ denotes the set $\pi_j(I)$. $\bar{U}_n$ belongs to $\tau_{box}$, so the sets $U_n:=\bar{U}_n \cap \mathbb{R}^{\infty}$ belong to $\tau_{sub}$. Therefore $U_n\in \tau_{col}$ $\forall$ $n\in \mathbb{N}$. Note that for each $n$, only a finite number of the elements $x_l$ belong to $U_n$.

Clearly, $\lbrace U_n \rbrace_{n\geq 1} $ is an open cover of $C$, and since $C$ is compact we can find a large enough $N\in \mathbb{N}$ such that $C\subseteq \bigcup_{n=1}^N U_n=U_N$, but there is still an infinite number elements of our sequence $\lbrace x_l\rbrace\subseteq C$ not in $U_N$, which is a contradiction. This finishes the proof.