# Why are there no filtered limits?

I just read the Wikipedia-article about filtered category, and now I wonder why it mentions filtered colimits and cofiltered limits, but not filtered limits. On the nLab-article titled filtered limit a filtered limit is even defined as a limit over a cofiltered category.

Is there a reason for this? The first thing that comes to my mind when I try to think of a limit over a filtered (but not cofiltered) category is a pullback. So apart from this simple case, are limits over filtered categories not very useful, or well-behaved, or whatever?

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Filtered colimits have special properties in many concrete categories of interest, and cofiltered limits are their formal dual. I do not know of any good properties of filtered limits. – Zhen Lin Jul 16 '13 at 0:43
The special property Zhen alludes to is that in $\text{Set}$, filtered colimits are precisely the ones which commute with finite limits. This implies various other nice things. There is no dual property of filtered limits, to my knowledge. – Qiaochu Yuan Jul 16 '13 at 0:47

If a category $C$ has a terminal object, then, as far as limits in $C$ are concerned, "filtered" would be useless information. To see this, consider any functor into $C$, say $F:I\to C$. Define $I^+$ to be the category obtained by adjoining a terminal object to $I$. (If $I$ already has a terminal object, ignore it and adjoin a new one.) So the objects of $I^+$ are those of $I$ plus the new terminal object $1$, and the morphisms of $I^+$ are those of $I$ and, for each object of $I^+$, a single morphism from that object to $1$. Extend $F$ to a functor $F^+:I^+\to C$ by sending the new $1$ of $I^+$ to the terminal object of $C$ and defining $F$ on the new morphisms of $I^+$ in the only possible way. Then, unless I've stupidly overlooked something, $I^+$ is filtered, and the limit of $F^+$ agrees with that of $F$. In other words, any limit at all (like that of $F$) can be turned into a filtered limit (like that of $F^+$) by a trivial modification.