Compactness of a set in order topology Consider a partial order in $R^2$ 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 $R^2$ defined by the above order, how can I conclude that $[0,1]$×$[0,1]$ is not compact?
My thought:I have two doubts.


*

*I am not getting how the given relation < is a partial ordering on $R^2$ when it doesn't seem to be reflexive to me. However, it is antisymmetric and transitive.

*If I agree that the relation < is a partial ordering then am I right to say that all the points of unit square [0,1]×[0,1] belong to the order topology of $R^2$? Then how it is not compact?
Please tell me at what point I am wrong and help me to reach the result. Thanks in advance.


Hint: $[0,1]^2$ is a subset of $\mathbb R^2$, so it is given a subspace topology (coming from the order topology of $\mathbb R^2$). Note that in this topology, $\{x\} \times [0,1]$ is open as 
$$\{x\} \times [0,1] = [0,1]^2 \cap\big( \{x\} \times (-0.1,1.1)\big)$$
and $\{x\} \times (-0.1,1.1)$ is open in the order topology. Then the subset of open sets
$$\{ \{x\}\times [0,1]| x\in [0,1]\}$$
is an open cover of $[0,1]\times [0,1]$. But this open cover has no finite subcover. 
