Compactness of subspaces of $\mathbb R^\infty$. Let $\mathbb R^\infty=\bigcup \mathbb R^i$ (identifying $\mathbb R^i$ with subspace of $\mathbb R^{i+1}$) and $U\subset \mathbb R^\infty $ is open iff $U\cap \mathbb R^i$ is open in $\mathbb R^i$ for every $i$. How can i show that every compact subset of $\mathbb R^\infty$ is contained in $\mathbb R^i$ for some $i$.
 A: We can identify $\mathbb R^\infty$ with the set of sequences $x\colon\mathbb N\to \mathbb R$ with finite support (and have $x\in\mathbb R^i$ iff $x(\nu)=0$ for all $\nu>i$).
Let $K\subseteq \mathbb R^\infty$ be compact.
The sets $$U_r=\{\,x \in\mathbb R^\infty\mid \forall n\colon |x(n)|<r\,\}$$ are open and $\bigcup_{r>0} U_r=\mathbb R^\infty$, hence by compactness $K\subseteq U_R$ for some $R>0$ (using the fact that $r<s$ implies  $U_r\subseteq U_s$).
For $n\in\mathbb N$, let $r(n)=\sup_{x\in K}|x(n)|$. Then $0\le r(n)\le R$. 
Assume there exists a strictly increasing sequence $\{n_k\}_{k\in\mathbb N}$  of naturals with $r(n_k)>0$ for all $k\in\mathbb N$.
Let $$V_k=\{\,x\in \mathbb R^\infty\mid \forall j>k\colon |x(n_j)|<\tfrac12 r(n_j)\,\}.$$
Then $V_k$ is open and $K\subseteq \bigcup_{k\in\mathbb N} V_k$.
Then by compactness $K\subseteq V_m$ for some $m$ (again, using the fact that  $k<m$ implies $V_k\subseteq V_m$).
But then for $j>m$ we have $|x(n_j)|<\frac12 r(n_j)$ for all $x\in K$, contradicting $0<r(n_j)=\sup_{x\in K}|x(n_j)|<\infty$.
Therefore the assumed existence of a sequence $\{n_k\}_{k\in\mathbb N}$ as above is false, i.e. there exists $i\in\mathbb N$ with $r(\nu)=0$ for all $\nu>i$. In other words, $K\subseteq \mathbb R^i$. $_\square$
A: $\mathbb{R}^{\infty} \simeq colim_{i \in \mathbb{N}}\mathbb{R^i}$, where I am implicitly considering the obvious functor $F:\mathbb{N} \to Top$ which acts as $i \mapsto \mathbb{R}^i$. Each map $\mathbb{R}^i \to \mathbb{R}^{i+1}$ is a closed $T_1$ inclusion, meaning that it's injective and $\forall U \subset \mathbb{R}^i$ open subset $\exists V\subset \mathbb{R}^{i+1}$ such that the preimage of $V$ is $U$, moreover each point out of the image is closed. Since each compact set is finite relative to closed $T_1$ inclusions, i.e. we have the isomorphism $colim_{i \in \mathbb{N}} Top(K,\mathbb{R}^i)\simeq Top(K,colim_{i \in \mathbb{N}} \mathbb{R}^i)\simeq Top(K,\mathbb{R}^{\infty})$ (Hovey's "Model Categories" Prop. 2.4.2), you get the desired factorization.
A: Suppose that $K\subset\mathbb R^\infty$ is not contained in any $\mathbb R^i$. Then choose a sequence $(a_n)$ in $K$ such that $a_n\notin\mathbb R^n$ and set $A:=\{x_n\}$. Let $B$ be an arbitrary subset of $A$. The set $B$ contains only finitely many points of every $\mathbb R^i$, hence its preimage in $\mathbb R^i$ is closed, hence $B$ is closed in $\mathbb R^\infty$. This shows that $A$ is a closed discrete subset of $K$. Since $A$ is discrete and infinite, it is not compact. Since $A$ is closed in $K$ and $K$ hausdorff (you will have to show this), $K$ cannot be compact either.
