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# order topology

Let $(X,\leq)$ be a linearly ordered set. The *order topology* on $X$ is defined to be the topology $\mathcal{T}$ generated by the subbasis consisting of open rays, that is sets of the form

$(x,\infty)=\{y\in X|y>x\}$ |

$(-\infty,x)=\{y\in X|y<x\},$ |

for some $x\in X$.

This is equivalent to saying that $\mathcal{T}$ is generated by the basis of open intervals; that is, the open rays as defined above, together with sets of the form

$(x,y)=\{z\in X|x<z<y\}$ |

for some $x,y\in X$.

The standard topologies on $\mathbb{R}$, $\mathbb{Q}$ and $\mathbb{N}$ are the same as the order topologies on these sets.

If $Y$ is a subset of $X$, then $Y$ is a linearly ordered set under the induced order from $X$. Therefore, $Y$ has an order topology $\mathcal{S}$ defined by this ordering, the *induced order topology*. Moreover, $Y$ has a subspace topology $\mathcal{T}^{{\prime}}$ which it inherits as a subspace of the topological space $X$. The subspace topology is always finer than the induced order topology, but they are not in general the same.

For example, consider the subset $Y=\{-1\}\cup\{\frac{1}{n}\mid n\in\mathbb{N}\}\subseteq\mathbb{Q}$. Under the subspace topology, the singleton set $\{-1\}$ is open in $Y$, but under the order topology on $Y$, any open set containing $-1$ must contain all but finitely many members of the space.

A chain $X$ under the order topology is Hausdorff: pick any two distinct points $x,y\in X$; without loss of generality, say $x<y$. If there is a $z$ such that $x<z<y$, then $(-\infty,z)$ and $(z,\infty)$ are disjoint open sets separating $x$ and $y$. If no $z$ were between $x$ and $y$, then $(-\infty,y)$ and $(x,\infty)$ are disjoint open sets separating $x$ and $y$.

## Mathematics Subject Classification

54B99*no label found*06F30

*no label found*

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