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I found that the x-axis in the Moore plane is a closed and discrete set and the real line with the lower limit Topology don't have a set like that but I don't know how to conclude that there is not an homeomorphism between the Moore plane and the real line with the lower limit Topology. Thanks for any help.

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You know that the Moore plane has some topological property that the lower limit topology doesn't have. Use this and the fact that homemorphisms preserve topological properties to complete the proof.

Suppose $X$ is the Moore plane and $Y$ is the real line with the lower limit topology, and suppose by contradiction that $f:X\to Y$ is a homeomorphism.

Let $D\subseteq X$ be the real line in the Moore plane, which is a closed uncountable discrete subspace. Since $f$ is a homeomorphism, the restriction $g:D\to f(D)$ given by $g(d)=f(d)$ is also a homeomorphism. (Verify this.) Thus $f(D)$ is a closed uncountable discrete subset of $Y$. Since we know that $Y$ doesn't have a subspace like this (because, for example, it is Lindelöf), then we are done.

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  • $\begingroup$ Thanks so much! $\endgroup$ – Gonzalo Castillo Oct 15 '17 at 2:35

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