Hi all,
Here I have two problems that really I can´t know how solve.
For $X = S^2 \cup \{x_0\}$ where $x_0 \in \mathbb{R}^3 \setminus S^2$, being $S^2$ the sphere. In $X$ we consider the usual topology for points in $S^2$ and for $x_0$ the open sets are $(U \setminus (0,0,1)) \cup \{x_0\}$, being $U$ a neighborhood of $(0,0,1)$ in $S^2$.
Consider $X = \mathbb{R}^2 \times \mathbb{R}$, where in $\mathbb{R}^2$ we take the usual topology and in $\mathbb{R}$ the discrete topology (everything is clopen).
In both situations (1. and 2.) prove that any point of $X$ has a neighborhood homemorphic to an open set of $\mathbb{R}^2$ but $X$ is not a surface.
I suppose that in 1. the problem is with the open neighborhood of $x_0$ and in 2. is with the discrete topology. But any hint more detailed? I know that there are so questions, but I´ve not any idea about how you can do it all.
Thank you very much for the support!
Supposing surface means 2-dimensional topological manifold:
(1) Your "surface" isn't Hausddorff.
(2) Your "surface" isn't second-countable.