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I have to show the following statement: enter image description here

My idea is to say that if $(X,\tau)$ is a topological space locally homeomorphic to $\mathbb{R^n}$, then $\forall x \in X$ there exist an open neighborhood $U$ such that $f$ maps U homeomorphically onto an open subspace $V$ of $\mathbb{R^n}$ with the euclidean topology.
Is it sufficient to say that this subspace can be the same $\mathbb{R^n}$ and thus, as $\mathbb{R^n}$ is homeomorphic to an open ball centered at zero, then (using the transitive property) this neighborhood is homeomorphic to an open ball centered at zero? This would prove the statement, but I am not sure what I am saying is right.

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  • $\begingroup$ This sounds like a tautology. Please, provide the definition of "locally Euclidean" that you are using. $\endgroup$ – Moishe Kohan Feb 27 '17 at 16:33
  • $\begingroup$ A topological space $(X,\tau)$ is said to be locally euclidean if there exists a positive integer $n$ such that each point $x \in X$ has an open neighbourhood homeomorphic to an open ball about $0$ in $\mathbb{R^n}$. $\endgroup$ – math4everyone Feb 27 '17 at 19:21
  • $\begingroup$ Then why isn't your question is about a tautology? $\endgroup$ – Moishe Kohan Feb 27 '17 at 19:33
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I see what Moishe Cohen is saying. Basically all you are doing is re-writing the definition. But what you are saying is on the right track but looks a little vague. Can you make an attempt at the proof? It should be easier to prove in this form.

A Topological Space X is locally Euclidean $\Leftrightarrow$ every point of X has a neighbourhood homeomorphic to $\mathbb{R}^n$

$\Rightarrow$ is trivially true by definition.

you need to prove $\Leftarrow$

Hint: You need to use the definition of an open ball together with your homeomorphism

Soln: $\Leftarrow$:

Recall-locally homeomorphic to $\mathbb{R}^n$ means each $\mathbf p$ $\in$ $\mathbf X$ has an open neighbourhood $\mathbf U$ and $\exists$ a homeomorphism $\phi$:$\mathbf U$$\to$$\mathbf V$, $\mathbf V$ $\in$ $\mathbb{R}^n$ open.

Given any $\mathbf v$ $\in$ $\mathbf X$, let $\mathbf U$ be an open neighbourhood with $\phi$:$\mathbf U$$\to$$\mathbf V$ where $\mathbf V$ is an open subset of $\mathbb{R}^n$. Since $\mathbf V$ open, $\exists$ some open ball $\mathbf B$ around $\phi$($\mathbf q$) contained in $\mathbf V$ and ${\phi}^-1$($\mathbf B$) is a neighbourhood of $\mathbf q$ homeomorphic to an open ball.

Since every open ball in $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$ itself the result has been proved. $\square$

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