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I'm reviewing past papers for a topology exam, and I can't answer this question on topological manifolds:

Let $$X= \{(x,-1):x\in \mathbb{R}\}\cup\{(\cos \theta, \sin \theta ): 0 \le \theta \le 2\pi \}.$$

Show that $X$ s not a topological 1-manifold.

Clearly $X$ is Hausdorff and has a countable base, I guess there's no open cover to $X$ of sets homeomorphic to open subsets of $\mathbb{R}$, but how can I show this?

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You'd approach this in a different way than you're thinking of. Consider the intersection $U$ of $X$ with any small open ball around $(0,−1)$. Remove the point $(0,−1)$ from $U$: the result has 4 connected components. If $U$ were homeomorphic to an open interval $I$ in $\mathbb{R}$, then removing the corresponding point from $I$ would disconnect the interval into 4 components; but removing a point from an interval disconnects it into only 2 components. Homeomorphisms preserve the number of connected components.

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  • $\begingroup$ That makes sense, thanks! $\endgroup$ – James Nov 8 '15 at 8:47

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