Question on what constitutes surfaces in R3

Question

Our lecturer gave us the following definition of a surface in $\mathbb{R}^3$: $\Gamma$ Is a surface in $\mathbb{R}^3$ if for all $y\in \Gamma$ there exists a coordinate patch $\sigma : D\subset\mathbb{R}^2 \rightarrow \mathbb{R}^3$ such that $y \in \text{Im}(\sigma)\subset\Gamma$. From this definition, is it not inferred that the union of two surfaces is a surface? I am confused because he went on to mention that the union of the boundary of the unit sphere and the set $\{(x,y,0)|x^2+y^2<1\}$ is not a surface. Clearly these are both surfaces so why isn’t their union?

Answer 1

Let $S$ be the unit sphere $\{(x, y, z) \mid x^2+y^2+z^2=1\}$ and $T = \{(x, y, 0) \mid x^2 +y^2<1\}$ be the two surfaces you mention and let $\Gamma = S \cup T$. Then the point $e = (1, 0, 0)$ in $\Gamma$ has no neighbourhood in $U$ that is homeomorphic with an open set $D$ of $\Bbb{R}^2$. To prove this takes a little work, but it should be clear visually: any neighbourhood of $e$ in $\Gamma$ looks locally like the union along their edges of three half-planes and no open subset of $\Bbb{R}^2$ looks like that.