a) Show that $[0,1]$ is not compacts as a subspaces of $R_K$
i know that $R_K $ is finer then R since its basis contain the basis of $R$
$R_K =(a,b) - \frac{1}{n}$ which is not open because $\frac{1}{n} $ is not closed as complement of open set is closed
Recall that the K-topology is generated by the basis {$(a, b), (a, b) − K | a < b$}, where
K = { $\frac{1}{n}| n\in Z^+$} for each i ,let $U_i = (\frac{1}{i} ,2) U (-1,1) - k:$ the open cover {${u_i}$} of $[0,1]$ does not have finite subcover so$ [0,1]$ is not compacts
A topology strictly finer than a compact Hausdorff topology is not compact.
Indeed if $K$ is compact and $X$ is Hausdorff then every continuous bijection from $K$ to $X$ is a homeomorphism (since closed).
Fact: In a compact space $X$, every countably infinite subset $A$ has an $\omega$-accumulation point $p\in X$: i.e. a point $p$ such that every (open) neighbourhood of $p$ contains infinitely many points of $A$.
Proof: suppose not, then every $x \in X$ would have an open neighbourhood $O_x$ such that $O_x \cap A$ is finite. Taking a finite subcover by compactness, we easily see that this would imply that $A$ is finite, contrary to how $A$ is chosen. Contradiction.
In $[0,1]$ in the $K$-topology this fails for $A=K$, as for any $x \notin K$, $(x-1,x+1)\setminus K$ is a neighbourhood of $x$ in this topology that has no points of $K$. And each $x \in K$ has a small neighbourhood that misses all other points of $K$ ($K$ is discrete in itself). So no $x$ can be an $\omega$-accumulation point of $K$.
