Remark
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[0HY] Let \((J,≤)\) be a non-empty set with filtering order. We know from [06V] that \(J\) has no maximum. We extend \((J,≤)\) by adding a point ”\(∞\)”: Let’s set \(I=J∪ \{ ∞\} \) and decide that \(x≤ ∞\) for every \(x∈ J\). It is easy to verify that \((I,≤)\) is a direct order, and obviously \(∞\) is the maximum \(I\). 1 Let \(𝜏\) be the topology defined in [0HW]. We know that \(∞\) is an accumulation point. This topology can explain, in a topological sense, the limit already defined in [0FR], and other examples that we will see in Sec. [2B8].