- E14
[2F5] Consider a totally ordered set \(X\) (that has at least two elements), and the family \(\mathcal F\) of all open-ended intervals
\begin{align} (x,β) {\stackrel{.}{=}}\{ zβ X : x{\lt}z\} ~ ~ ,~ ~ (-β,y){\stackrel{.}{=}}\{ zβ X : z{\lt}y\} ~ ~ ,\nonumber \\ ~ ~ (x,y){\stackrel{.}{=}}\{ zβ X : x{\lt}z{\lt}y\} \label{eq:intervalli_ topologia_ ordine} \end{align}for all \(x,yβ X\).Β (Cf. [07D].) Prove that this is a base for a topology, i.e. that it satisfies [0KZ]. So \(\mathcal F\) is a base for the topology \(π\) that it generates. This topology \(π\) is called order topology.
If \(X\) has no maximum and no minimum, then only the intervals \((x,y)\) are needed to form a base for \(π\). This is the case for the standard topologies on \(β\), \(β\), \(β€\),
EDB β 2F5
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English
Authors:
"Mennucci , Andrea C. G."
.
Bibliography
Book index
Book index
- space, topological
- topological space
- base, (topology)
- order topology
- topology, order β
- order, total
- base, (topology)
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