# Subspace of Lindelöf space is not Lindelöf: Example

The Munkres' topology book provides Example 30.5 (p.193, 2nd Ed) for a subspace of a Lindelöf space that need not be Lindelöf as follows:

The ordered square $I_0^2$ is compact; therefore it is Lindelöf trivially. However, the subspace $A = I \times (0,1)$ is not Lindelöf. For $A$ is the union of the disjoint sets $U_x = \{x\} \times (0,1)$, each of which is open in $A$. This collection of sets is uncountable, and no proper subcollection covers $A$.

I could largely understand this argument. But it is not clear to me how $U_x$ is open in $A$. Could someone please explain? Thanks.

RD

• If you want a simpler example, let $X$ be any set of uncountable cardinality, and let $Y = X \cup \{x\}$ for some $x\notin X$. Declare a set to be open iff it's $Y$, empty, or a subset of $X$. Then $Y$ is compact (the only open set which contains $x$ is $Y$ itself), but the subspace $X$ is not Lindelof. – Jason DeVito May 5 '15 at 4:54

The set $U_x$ is open, since the topology on the ordered square is not the standard Euclidean topology. The definition of the order on the square is lexicographic, ie $(x_1,y_1) < (x_2, y_2)$ if either $x_1 < x_2$ or if $x_1 = x_2$ and $y_1 < y_2$. The corresponding order topology is generated by sets of the form $$\{ (x, y) : (x_1, y_1) < (x,y) < (x_2, y_2)\}$$
In this example $U_x$ is obtained by taking $(x_1, y_1) = (x,0)$ and $(x_2, y_2) = (x,1)$.
A simpler example is, in Munkres' notation, $\overline{S_{\Omega}}$. (The closure of the first uncountable well ordered set in the order topology.)
The space is compact and therefore Lindelöf trivially, (It is compact because it is a well ordered set with a maximum in the order topology, but you can find other ways to prove it's Lindelöf.) but $S_{\Omega}$, the same space without the maximum, is not Lindelöf as the open cover $\{[0,\alpha)\}_{\alpha\in S_{\Omega}}$, has no countable subcover. Note every countable set in $S_{\Omega}$ is bounded.