Why isn't the union of the coordinate axes a manifold, using the formal definition of a manifold?

Question

Let me call $M$ a manifold iff locally it is given as the graph of a $C^1$ function. Then without appealing to "remove a point" type arguments, why isn't the union of the $x$ and $y$ coordinate axes (the solution set of $xy=0$) a manifold? That is, why isn't this locally a graph around the origin?

Answer 1

The formal definition of manifold I know is that it's a topological space that's locally homeomorphic to $\mathbb R^N$. Thus in order for the union of the axes to be a 1-manifold, there must be an open neighborhood (in the putative manifold) of $\langle 0,0\rangle$ that is homeomorphic to $\mathbb R$.

By definition, every such open neighborhood contains a ball $B_{2\delta}(\langle0,0\rangle)$ for some $\delta>0$, and therefore it contains the points $P\langle0,\delta\rangle$, $Q\langle-\delta,0\rangle$ and $R\langle\delta,0\rangle$.

These three points have the property that there's a path from $P$ to $Q$ not passing through $R$, a path from $Q$ to $R$ not passing through $P$, and a path from $R$ to $P$ not passing through $Q$. But there's no triple of points with this property in $\mathbb R$, so the neighborhood of $\langle 0,0\rangle$ cannot possibly be homeomorphic to $\mathbb R$.

The intersecting lines cannot be a manifold of higher dimension either, because every neighborhood of $\langle 0,0\rangle$ has an open subset that is homeomorphic to an interval, and $\mathbb R^d$ doesn't when $d\ge 2$.

Answer 2

A function is single-valued, but in any neighborhood of $0$ we have an infinite number of points of the form $(0,y)$ for different $y$.