Definition of the Space $\mathbb{R}^\infty$ I'm trying to understand the definition of the topological space $\mathbb{R}^{\infty}$, which Hatcher defines as $\cup_n \mathbb{R}^n$ in his book Algebraic Topology. I'm having trouble making sense of this union since $\mathbb{R}^m$ is not literally a subset of $\mathbb{R}^n$ for $m < n$. Certainly the first embeds into the second, but in order to form the space $\bigcup_n \mathbb{R}^n$ and give it the weak topology we need an infinite chain of subsets $\mathbb{R}^0 \subset \mathbb{R}^1 \subset \mathbb{R}^2 \subset \cdots$. We can almost obtain this by replacing each Euclidean space by its embedded image in higher-dimensional Euclidean spaces, but this approach falls short because it will only give us finite subset chains. How do we get around this formal problem?
 A: The way to get around this is a classic "trick". The $\mathbb{R}^n$ don't all appear as embedded subspaces of any $\mathbb{R}^n$, so we can't just take a union of them with the subspace topology. However, by design all the $\mathbb{R}^n$ appear as subsets of $\bigsqcup\limits_{n} \mathbb{R}^n$. Then we want to force the canonical inclusion $\mathbb{R}^n \rightarrow \mathbb{R}^{n+1}$ to be one of subspaces by using the quotient topology on the relation that identifies the path component $\mathbb{R}^n$ with its image in the path component $\mathbb{R}^{n+1}$ under the canonical inclusion.
In other words the space $\mathbb{R}^\infty$ is $\bigsqcup\limits_{n} \mathbb{R}^n / \sim$ where $(x_1, \dots, x_k) \sim (x_1, \dots, x_k,0)$ for all k.
This is the usual model of the colimit of $\mathbb{R} \rightarrow \mathbb{R}^2 \rightarrow \dots$ that the other user mentions in their answer.
A: Let $X$ be the space of all real sequences $(x_n)_\mathbb{N}$ with the property that there is $N\in\mathbb{N}$ such that $x_n=0$ if $n>N$. i.e. all but finitely many terms are nonzero. Such sequences are said to be eventually zero. Then $X$ is a subset of the set of all sequences $\mathbb{R}^\mathbb{N}$.
For any $k\geq0$ there is an inclusion $j_k:\mathbb{R}^k\rightarrow X$ which identifies $(y_1,\dots,y_k)\in\mathbb{R}^n$ with the sequence
$$x_n=\begin{cases}y_n&1\leq n\leq k\\0&\text{otherwise}. \end{cases}$$
($\mathbb{R}^0$ identifies with the sequence which is constant at $0$.) Then any point $\alpha\in X$ is in the image of some $j_k$. Give $X$ the weak topology with respect to the inclusions $j_k$, $k\geq0$. Then $X$ is $\mathbb{R}^\infty$ (up to homeomorphism).
Each $\mathbb{R}^n$ embeds in $X$ as a closed subspace. Each point of $X$ is contained in some $\mathbb{R}^n$ (or if you prefer, the image of some $j_n$). The image of $j_n$ is a subset of the image of $j_{n+1}$.
If we identify $\mathbb{R}^n$ with its image in $X$ then we have an infinite sequence of spaces
$$\mathbb{R}^0\subseteq \mathbb{R}^1\subset \mathbb{R}^2\subset\dots\subset \mathbb{R}^n\subset \dots$$
If we are not willing to identify $\mathbb{R}^n$ with its image in $X$, then we still have an infinite sequence of subspaces
$$j_0(\mathbb{R}^0)\subseteq j_1(\mathbb{R}^1)\subset j_2(\mathbb{R}^2)\subset\dots\subset j_n(\mathbb{R}^n)\subset \dots$$
A: I think this is just a common shorthand in algebraic topology. More rigorously, I think he's defining $\mathbb{R}^\infty$ as the colimit of the diagram $\mathbb{R}^1\hookrightarrow\mathbb{R}^2\hookrightarrow\dots$ I believe the topology this colimit inherits from each Euclidean space will be the one you're thinking of, but I could be wrong.
