Basic question on the topologies associated with the real lines Let $X = \mathbb{R}$. The standard topology is $\mathcal{T}$ in where the open sets are just elements of the form $(a,b)$. Let $\mathcal{T}_L$ be the topology where the open sets are $[a,b)$. Next, let $K = \{ \frac{1}{n} : n \in \mathbb{N} \} $. Let $\mathcal{T}_K$ be the topology where open sets are of the form $(a,b) \setminus K $ with the collection of all intervals of the form $(a,b)$
.
Problem:
The $L$-Topology and the $K-$topology are strictly finer than the standard topology.
Attempt:
So, I am trying to show that $\mathcal{T} \subset \mathcal{T}_L$ and $\mathcal{T} \subset \mathcal{T}_K$. Take an element in the standard topology, say $(a - \frac{1}{n}, b ) $. Then we have the intersection of all of the them is again in $\mathcal{T}$
$$ \bigcap_{n \geq 1} (a - \frac{1}{n}, b) = [a,b) \in \mathcal{T}_L $$
Hence, $\mathcal{T} \subset \mathcal{T}_L $
Is this a correct approach? How Can I tackle the second inclusion? thanks for any help.
 A: First, the elements of $\mathcal{T}$ are arbitrary unions of open intervals: $\mathscr{B}=\{(a,b):a,b\in\Bbb R\text{ and }a<b\}$ is a base for $\mathcal{T}$, but it’s not the topology itself. Similarly, $\mathscr{B}_{\mathcal{L}}=\{[a,b):a,b\in\Bbb R\text{ and }a<b\}$ is a base for the topology $\mathcal{T}_{\mathcal{L}}$, but it’s not the whole topology. For instance,
$$(0,1)=\bigcup_{n\ge 2}\left[\frac1n,1\right)$$
is in $\mathcal{T}_{\mathcal{L}}\setminus\mathscr{B}_{\mathcal{L}}$. Finally, 
$$\mathscr{B}_{\mathcal{K}}=\{(a,b):a,b\in\Bbb R\text{ and }a<b\}\cup\{(a,b)\setminus K:a,b\in\Bbb R\text{ and }a<b\}$$
is a base for $\mathcal{T}_{\mathcal{K}}$, but it’s not the whole topology. (Note: I’m assuming that $\mathcal{T}_{\mathcal{K}}$ is supposed to be the K-topology.)
Note that $\mathscr{B}\subseteq\mathscr{B}_{\mathcal{K}}$; from this you can easily show that $\mathcal{T}\subseteq\mathcal{T}_{\mathcal{K}}$, since any set that is a union of members of $\mathscr{B}$ is obviously a union of members of $\mathscr{B}_{\mathcal{K}}$. 
To show that $\mathcal{T}\subseteq\mathcal{T}_{\mathcal{L}}$, it suffices to show that $\mathscr{B}\subseteq\mathcal{T}_{\mathcal{L}}$; can you see why? And you can prove that by adapting the argument that I used above to show that $(0,1)\in\mathcal{T}_{\mathcal{L}}$.
Now all that remains is to find a set in $\mathcal{T}_{\mathcal{L}}$ that’s not in $\mathcal{T}$, and a set in $\mathcal{T}_{\mathcal{K}}$ that’s not in $\mathcal{T}$. 


*

*For the former, any member of $\mathscr{B}_{\mathcal{L}}$ will work; for simplicity you might as well take $[0,1)$. The proof that it’s not open in the usual topology is very much like what we did in this question.

*For the latter, try to show that $(-1,1)\setminus K$ is not open in the usual topology. A sketch is likely to be helpful; can you find a point of the set that is not in its Euclidean interior?
A: Of course, in all three cases we assume, that the given set is the basis of topology. Let us observe, that only finite unions of open sets should be open, hence your attempt is not correct.
It is known (or easy to show), that every open set of reals is the countable union of disjoint open intervals. But
$$
\bigcup_{n=n_0}^{\infty}\left[a+\frac1n,b\right)=(a,b),
$$
hence every open set of $\cal T$ is generated by elements of ${\cal T}_L$, but intervals of the form $[a,b)$ are not open in $\cal T$, hence ${\cal T}_L$ is strictly stronger than $\cal T$.
We can see, that ${\cal T}_K$ generates all open sets of the topology on $\mathbb{R}\setminus K$ generated by $\cal T$ and only them. For example point $1/2$ is not an elemet of any open (${\cal T}_K$) set. Hence ${\cal T}_K$ is strictly weaker than $\cal T$.
