is $[0,1]$ compact in topology generated by $\{(a, b): aI have to prove that, mi idea is create a coberture especific, and show that i can't take a finite collection, i was trying to see which can function, i was trying to check $[0, 1]$ in $k-$ topology to have an idea, but i did'not get it.
My teacher suggested me that i could take an open set with the form $(a, b)\cap \mathbb{Q}$ to cover rational numbers in $[0,1]$ and then take a sequence with rational numbers such that converges to $1$, and then create open intervals such that cover all the irrational numbers in $[0, 1]$, but i'm very stuck, i will really appreciate any clue.
 A: Hint: As in the suggestion, consider the covering $$[0,1]\subseteq \left((-1,2)\cap\mathbb Q\right)\cup\bigcup_{n=1}^{\infty}\left(0,1-\frac{1}{n}\right).$$
A: There are different ways to disprove compactness. Denote by $[0,1]_k$ the interval with the topology under scrutiny and by $[0,1]_u$ the interval with the usual topology.
Clearly the topology under scrutiny is finer than the usual topology, so the identity function $[0,1]_k\to[0,1]_u$ is continuous.
Suppose $[0,1]_k$ is compact. If a subset $A$ is closed, then it is compact. Then its image under a continuous function is compact. Since $[0,1]_u$ is Hausdorff, compactness implies closedness.
Now observe that the irrational numbers in $[0,1]_k$ are a closed set.
More generally, if $(X,\tau)$ is compact and Hausdorff, then $(X,\sigma)$ is not compact under any Hausdorff topology $\sigma$ such that $\sigma\subsetneq\tau$.
A: Let $\Bbb P = [0,1]\setminus \Bbb Q$, the irrationals in $[0,1]$.
If $[0,1]$ in this $k$-topology were compact, then $\Bbb P$ would also be, but  $\Bbb P$ has the same topology as subspace of the $k$-topology as it has in the standard topology on $[0,1]$ and $\Bbb P$ is dense in $[0,1]$ (usual ) so certainly not compact. So $[0,1]$ is not compact in the $k$-topology.
