# Union of two self-intersecting planes is not a surface

I need to show that the union of xy-plane and xz-plane, i.e. the set $S:=\lbrace (x,y,z)\in\mathbb{R}^3 : z=0 \mbox{ or } y=0\rbrace$, is not a surface. Here is my claim,

$\textbf{Claim :}$ Suppose $p$ is the point $(0,0,0)$ and $U:=S\cap B(p,\epsilon)$, where $\epsilon > 0$. Then $U$ cannot be homeomorphic to any open set of $\mathbb{R}^2$.

I need help in proving the above claim. Here are my thoughts: Suppose $U$ is homeomorphic to an open set $V$ of $\mathbb{R}^2$ via homeomorphism $f$.

Now if we remove the $x$-axis from $S$ then it will have $4$ connected components, so if we can show that removal of the image of $x$-axis from $V$ can leave at most three connected components then we are through, is it the case?

Another approach : if we remove the point $p$ from $S$, then the fundamental group of the resulting space is same as the fundamental group of $X:=\lbrace (x,y,z)\in\mathbb{R}^3 : z=0 \mbox{ and } x^2+y^2=1,\mbox{ or } y=0\mbox{ and } x^2+z^2=1\rbrace$. Can the two fundamental groups $\pi_1(X)$ and $\pi_1(V\setminus \lbrace f^{-1}(p)\rbrace )$ be same? If $V$ were an open ball then I don’t think they are same (though I don’t know the proof), but $V$ is any arbitrary open set.

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How about this: if you remove an arc joining two points on the boundary of a disc, you disconnect the disc; but you can remove an arc joining two points on the boundary of $U$ without disconnecting $U$. –  Gerry Myerson Feb 4 '13 at 6:22
@Gerry Myerson: But that would disconnect the disc, it may not disconnnect the open set in which it is sitting and every open set need not be a disc. –  pritam Feb 4 '13 at 6:29
Regarding "but V is any arbitrary open set.", it's not arbitrary. Since $V$ is homeomorphic to $S$, it must be connected and simply connected, i.e. it is a ball. Minus a point, it is an annulus. So $\pi_1(V \setminus \{f^{-1}(p)\}) = \mathbb Z$. But $\pi_1(X)$ is something else (try to show this). –  Marek Feb 4 '13 at 7:47
Regarding 4 connected components, it's correct. But to prove this, you'd need something on the order of Jordan Curve theorem (to show that number of components outside a simple curve in $\mathbb R^2$ is at most two). –  Marek Feb 4 '13 at 7:53
Part of the definition of "surface" is that every point has a neighborhood homeomorphic to a disk. So it's enough to show that $U$ can't be. –  Gerry Myerson Feb 4 '13 at 9:24