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Let $T_1$ be the standard topology on $\mathbb{R}$, $T_2$ the $K$-topology on $\mathbb{R}$, where $K=\{\frac{1}{n}\mid n\in \mathbb{N}\}$ and $K$-topology is nothing more than the topology generated by basis elements of form $(a,b) \setminus K$. Let $T_3 = \{(-\infty,a) \mid a \in \mathbb{R}\}$.

Then establish the relation between them.

I know that the discrete topology is strictly finer than any other topology on $X$. Also $K$-topology is strictly finer than standard topology. What about other two topologies here. Are they comparable?

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    Note that the $K$-topology basis also contains regular open sets $(a,b)$. – 2017-02-22

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The $K-$topology $\mathcal{T}_2$ is strictly finer than that of the usual topology $\mathcal{T}_1$.

And the usual topology $\mathcal{T}_1$ is strictly finer that of the topology $\mathcal{T}_3$.