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Let $A,B$ be two points not in the real line.

Let $S=(\Bbb{R} \setminus \{0\}) \cup \{A,B\}$

We define a topology in S as follows:

On $(\Bbb{R} \setminus \{0\})$, use the subspace topology inherited from $\Bbb{R}$, with open intervals as a basis. A basis of neighborhoods at $A$ is the set $\{I_A(−c,d) | c,d > 0\}$ and similarly, a basis of neighborhoods at $B$ is the set $\{I_B(−c,d) | c,d > 0\}$, where $$I_A(−c,d)=(-c,0) \cup \{A\} \cup (0,d)$$ $$I_B(−c,d)=(-c,0) \cup \{B\} \cup (0,d)$$

I have to prove that this space is locally Euclideian,second countable but not Hausdorff. Thus the space is not a topological manifold.

I have proven that this space is locally euclideian but not Hausdorff.

For second countability i'm not sure if my argument is correct.

Second countability is a hereditary property so $\Bbb{R} \setminus \{0\}$ is second countable as a subspace of the second countable real line.

Thus exists a countablle basis $C=\{B_n|n \in \Bbb{N}\}$ for $\Bbb{R} \setminus \{0\}$

Now we take the sets $$D_A=\{I_A(c,d)|c,d>0 \text{ and } c,d \in \Bbb{Q}\}$$ $$D_B=\{I_B(c,d)|c,d>0 \text{ and } c,d \in \Bbb{Q}\}$$

Now we take the countable union $$W=C \cup D_A \cup D_B$$.

Is $W$ a the correct basis to prove the statement for second countability??

Thank you in advance.

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Looks good so far!

Another approach is to observe the space can be expressed as $S = X \cup Y$ where the subspaces $X$ and $Y$ are copies of the real line hence second countable. There should be a theorem somewhere that says the union of two second countable subspaces is itself second countable.

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