Is $\lim\limits(\prod_{i\leq n} K)\cong\cup_i(\prod_{j\leq i}K)$ true? Consider a family of $K-$vector spaces $\prod_{i\leq n}K\to\prod_{i\leq n+1}K$ by embedding where embedding is done by identifying the basis $(1,0,\dots,0)\to (1,0,\dots,0,0),...$ and similarly for the rest. 
$\textbf{Q:}$ Is $\lim\limits_n(\prod_{i\leq n} K)\cong\cup_i(\prod_{j\leq i}K)$ true? Should I conclude $\cup_i(\prod_{j\leq i}K)=\prod_N K$?
Note that   LHS means $\prod_{j\in N}K$ whereas every element of RHS arises from a finite sequence. Do I even have an infinite non-trivial sequence in RHS but it seems that all my sequence on RHS must terminate at finite position? However, I can identify the limit in set theoretical sense and then use $lim_n(\prod_{i\leq n}K)=\sqcup_{n\in N}(\prod_{i\leq n}K)/\sim$ where $\sim$ identifies element by the canonical embedding. What is wrong with my thought process?
 A: This answer is intended to sum up the discussion in the comments. With the intention of clearing up some misconceptions, a slightly more categorical analysis is given before arriving at the concrete example of the post.
Take $(V_i)_{i \geq 1}$ a sequence of vector spaces with $V_i \subset V_{i+1}$. This gives a diagram
$$
V_1 \subset V_2 \subset \dots V_n \subset \dots  \tag{$\star$}
$$
over the category $\omega = 1 \to 2 \to 3 \to \dots$ which can be thought of $\mathbb{N}$ as a poset: we have an arrow $n \to m$ iff $n \leq m$. Concretely, the diagram $(\star)$ is given by mapping $i$ to $Fi = V_i$ for each $i \geq 1$ and $n \to m$ which is $n \to n+1 \to \dots \to m$ as the composition of inclusions from each $V_i$ to the next, from $n$ to $m$. Now, having constructed a diagram, one may wonder whether $\lim F$ and $\operatorname{colim} F$ exist, and how they can be described.
Note that in the category $\omega$, we have an initial object (namely $1$). Is is an instructive exercise to prove that if a category $J$ has an initial object $i$ and $F : J \to \mathcal{C}$ is a diagram, then $\lim J = Fi$. Hence, for $(\star)$ we get
$$
\lim J = F(1) = V_1.
$$
As for the colimit of $(\star)$, I claim we have $\operatorname{colim} F \simeq \bigcup_{i \geq 1}V_i$. It suffices to see that the latter verifies the universal property of the colimit: we have inclusion maps $j_r : V_r \to \bigcup_{i \geq 1}V_i$ for each $s$ which commute with the inclusions $V_i \subset  \dots \subset V_{j}$. It suffices to check that they commute for $V_i \subset V_{i+1}$ (why?). Noting the inclusions $ s_k : V_k \hookrightarrow V_{k+1}$ we get that $j_{k+1}s_k(v) = j_{k}(v)$ for each $v \in V_k$. This is because the first expression is looking at $v$ as an element of $V_{k+1}$, and then embedding it in the union, and the latter is directly looking at $v$ as a vector on the union. Hence we have built a cone $(j_r : V_r \to \bigcup_{i \geq 1} V_i)_{r \geq 1}$ under $F$. Finally, let's prove that it is universal in this sense: if we have another cone $(\mu_r  : V_r \to W)_{r \geq 1}$, then the function
$$
\mu : v \in V_r \subset \bigcup_{i \geq 1}V_i \longmapsto \mu_r(v) \in W
$$
is well defined, linear and the only one such that the new cone factorizes via the original inclusions, i.e. $\mu$ is the only linear function such that $\mu j_r = \mu_r$. This completes the proof of $\operatorname{colim} F \simeq \bigcup_{i \geq 1} V_i$.
Now, going back to the original question: here $V_i = \mathbb{k}^i$ and so $\lim F  = \varinjlim V_i = \mathbb{k}$ and $\operatorname{colim} F = \varprojlim V_i = \bigcup_{i \geq 1} \mathbb{k}^i$. Hence if we interpret the limit as an inverse limit (i.e. a colimit) the statement is true. What is false is that $\bigcup_{i \geq 1} \mathbb{k}^i \simeq \mathbb{k}^{\mathbb{N}}$. The former has countable dimension, a basis given by $e_i = (0, \dots, 0, \overbrace{1}^i,0,\dots)$, but the latter does not.
