The union of $x$ and $y$ axis with subspace topology is not locally euclidean

Question

The union of $x$ axis and the $y$ axis is not manifold . It is Hausdorff and second countable, but it's not locally Euclidean. I was trying to prove it by contradiction, but I can't, the issue is supposed to be with the $(0, 0)$.

Possible duplicate of Union of $x$-axis and $y$-axis is not a manifold and this How to prove a cross is not locally homeomorphic to an euclidean space? — Kelvin Lois, Nov 08 '17 at 09:31
Can I apply the same argument if I want to proof the sets the union of the tow triangles <0,e1,e2> and <0,-e1,e2> it's not a closed manifold ?? — majduleen zeyadeh, Nov 08 '17 at 14:31

score 5 · Accepted Answer · edited Oct 29 '21 at 16:44

Take any neighbourhood of $(0, 0)$ in subspace topology. Suppose it homeomorphic to open interval $I \subset \mathbb{R}$ (as is the case with any other point except $(0, 0)$). Note that if you remove point $(0, 0)$ from the union of the $x$- and $y$-axes, you have four connected components and if you remove the corresponding point in $I$ you will have two connected components. By this you cannot have homeomorphism between them (by a property of continuous maps).

The union of $x$ and $y$ axis with subspace topology is not locally euclidean

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