What is the topology on the direct limit of $\Bbb R^i \rightarrow \Bbb R^j$ If we let $X$ be direct limit of 
$$\Bbb R \rightarrow \Bbb R^2 \rightarrow \Bbb R^3 \rightarrow \cdots $$ 
where each arrow is inclusion to first coordinates. 
What is a choice of $X$ (with a toplogy that coincides with the direct limit topology). 
EDIT: I believe it is subspace of $\Bbb R^{\Bbb N}$, sequence eventually $0$  taking values in $\Bbb R$, with the product topology.  I wonder if this is correct. 
 A: As mentioned in comments the direct limit is
$$L=\{(x_1,x_2,x_3,\ldots)\in\mathbb{R}^{\mathbb{N}}\ |\ x_i=0\text{ eventually}\}$$
The direct limit topology on $L$ is the topology coherent with $\{\mathbb{R}^n\ |\ n\in\mathbb{N}\}$ treated as subspaces of $L$ via obvious inclusions.
There's another natural choice for topology on $L$: the one induced by the Euclidean norm (note that the Euclidean norm is well defined on sequences that are eventually $0$).
I remember my surprise when I learned that these two are not the same. The example is as follows: let $v_i$ be a vector with $1/i$ on the $i$-th position and $0$ elsewhere. Then $E=\{e_i\}_{i=1}^\infty$ is closed in the direct limit (because $E\cap\mathbb{R}^n$ is finite for any $n$) but not closed in the Euclidean norm (because $\lVert v_i\rVert=1/i$ and so the sequence converges to $0$).
In fact it can be shown that the direct limit is not metrizable. In particular this also means that $L$ is not a subspace (even up to homeomorphism) of the product topology (at least a product of metrizable spaces) because countable product of metrizable spaces is metrizable.
