Does the Category of CPOs have omega^op limits?
Here's an attempt (please check!).

We have that $\bot_D = d$, where

$$
d_i = \bigsqcup^{D_i} \{\bot_i,f_i(\bot_{i+1}),f_i(f_{i+1}(\bot_{i+2})),\ldots\}
$$

By construction (and monotonicity), the above is the supremum of an $\omega$-chain inside $D_i$, so $d_i$ is well defined.

It is an element of $D$ since, by continuity,

$$
\begin{array}{ll}
&amp; f_i(d_{i+1}) \\
= &amp; f_i(\bigsqcup^{D_{i+1}} \{\bot_{i+1},f_{i+1}(\bot_{i+2}),f_{i+1}(f_{i+2}(\bot_{i+3})),\ldots\}) \\
= &amp; \bigsqcup^{D_{i}} \{f_i(\bot_{i+1}),f_i(f_{i+1}(\bot_{i+2})),f_i(f_{i+1}(f_{i+2}(\bot_{i+3}))),\ldots\} \\
= &amp; \bigsqcup^{D_{i}} \{\bot_i, f_i(\bot_{i+1}),f_i(f_{i+1}(\bot_{i+2})),f_i(f_{i+1}(f_{i+2}(\bot_{i+3}))),\ldots\} \\
= &amp; d_i
\end{array}
$$

To prove that $d$ is the bottom, we take any $x \in D$, and we observe

$$
\begin{array}{ll}
&amp; d_i \sqsubseteq x_i \\
\Leftarrow &amp; \forall k\geq i.\ f_i \circ \cdots \circ f_k(\bot_{k+1}) \sqsubseteq x_i \\
\Leftarrow &amp; \forall k\geq i.\ f_i \circ \cdots \circ f_k(\bot_{k+1}) \sqsubseteq f_i \circ \cdots \circ f_k (x_{k+1})\\
\Leftarrow &amp; \forall k\geq i.\ \bot_{k+1} \sqsubseteq x_{k+1}\\
\Leftarrow &amp; {\sf true}
\end{array}
$$