Recall that $\mathbb{R}_K$ denotes the real line with the $K$ topology. In other words, its basis elements are of the form $(a, b) – K$, where $K = \{1/n: n \in \mathbb{Z}_+\}$. On page 178 of Munkres's 'Topology' book, the author states that $(0, \infty)$ inherits its usual topology as a subspace of $\mathbb{R}_K$. But I disagree because $(0,1) – K$ is open in $\mathbb{R}_K$ and so its intersection with $(0,\infty)$ is open in $(0, \infty)$. In other words, $((0,1) – K) \cap (0, \infty) = (0,1) – K$ is open in $(0, \infty)$, but the standard topology on $(0, \infty)$ does NOT have $(0,1) – K$ as one of its open sets. So where am I wrong in my argument here? I am assuming that Munkres is correct and I am not. Thanks!
Does $(0, \infty)$ inherit its usual topology as a subspace of $\mathbb{R}_K$
general-topology
Related Solutions
You wrote:
We need to find an $A\times B$ open in $\mathbb{R}\times\mathbb{R}$ such that
This would be true if we were working with the product topology on $\mathbb R\times\mathbb R$. But in this example we are working with the order topology coming from the lexicographic order.
In this topology on $\mathbb R\times\mathbb R$ the set $\{1/2\}\times(1/2,3/2)$ is open, since it is precisely the set of all points which are between $(1/2,1/2)$ and $(1/2,3/2)$ (w.r.t. the linear order which we are working with).
The set $\{1/2\}\times(1/2,1]$ is not open in order topology (from the lexicographic order on $I\times I$) since every neighborhood of the point $(1/2,1)$ contains some points with $x$-coordinate greater than $1/2$. (Since the point $(1/2,1)$ does not have an immediate successor nor is it the greatest element in this order.)
The cover you referred to is (improved version) $\{(-2,2)-K\} \cup (\frac{1}{n},1)_{n \in N^+}$ which is an open cover of $[0,1]$ ($0$ and $1$ are covered by the first member, any $x>0, x < 1$ will be covered by some $(\frac{1}{n}, 1)$; and it's clear that all these sets are open in $\Bbb R_K$ by its definition.
Now if we have a finite subcover of it, we only use finitely many sets of the second type, so there is one with the largest $n$, say $\frac{1}{N}$ is that point. But then it's easy to find a point (or lots of points actually) that cannot be covered by this finite subcover, e.g. $\frac{1}{N+1}$ is one. (it's not covered by the first set, nor by any of the finitely many others by the choice of $N$.)
In $\Bbb R_K$ the sets $L = ( -\infty, 0)$ and $R = (0,+\infty)$ have the same topology as subspaces as they had in $\Bbb R$ in the usual topology. In particular, they are connected in $\Bbb R_K$ and both sets only add $0$ to their closure. So $\Bbb R_K = \overline{L} \cup \overline{R}$ is a union of two intersecting connected sets and thus connected.
Suppose there were a continuous path $p: [0,1] \to \Bbb R_K$ such that $p(0)=0, p(1)=1$. Then by connectedness of the image, $p[[0,1] \supseteq [0,1]$ and so $[0,1]$ is a closed subset of the compact image set, hence compact. But this contradicts what we’ve seen before that it is not compact. So $\Bbb R_K$ is not path-connected.
Best Answer
On the contrary, $(0,1)\setminus K$ is open in the standard topology: it is equal to the intersection of the open sets $(0,1)$ and $\Bbb R\setminus\big(\{0\}\cup K\big)$. The latter is open because $\{0\}\cup K$ is closed.