Consider the partial order in $\mathbb{R^2}$ given by the relation $(x_1,y_1) < (x_2,y_2)$ Either if $x_1 <x_2$ or if $x_1=x_2$ and $y_1 < y_2$ then in the order topology on $\mathbb{R}^2$ defined by the above order
choose the coorect statement
a) $[0,1] \times \{1\}$ is compact but $[0,1] \times[0,1]$ is not compact
b) $[0,1] \times [0,1]$ is compact but $[0,1] \times \{1\}$ is not compact
c) both $[0,1] \times [0,1]$ and $[0,1] \times \{1\}$ are compacts
d)both $[0,1] \times [0,1]$ and $[0,1] \times \{1\}$ are not compacts
My attempt : i got answer option $c)$
i was try visulalize the diagram i got option c that is both $[0,1] \times [0,1]$ and $[0,1] \times \{1\}$ are compacts, both are closed and bounded
Edit : according to henno sir
Is my logic is right or wrong pliz verified
Any hints/solution will be appreciated
thanks u
$[0,1] \times [0,1]$ fulfills the definition of a linear continuum (Munkres' style) so is compact (every subset has a supremum) but $[0,1] \times \{1\}$ is not compact. So b) is in fact correct.
$[0,1] \times \{1\}$ is homeomorphic to $[0,1]$ as a subset of the lower limit topology (using $(x,1) \to x$) and this is not compact as $1-\frac1n$ is a sequence without a cluster point, and it's a first countable space. Or use the cover $[0,x)$, $x < 1$ with $\{1\}$ that has no finite subcover.