Proving Topological Space L is Locally Homeomorphic to R but not Hausdorff

[AiRAVATA]

Hello guys. I need to prove the following:

Let $X=(\mathbb{R}\times \{0\})\cup(\mathbb{R}\times\{1\})$ and $(x,0)\sim (x,1)$ when $x \neq 0$. Prove that $L:=X/\sim$ is a topological space locally homeomorphic to $\mathbb{R}$, but is not Hausdorff.

In order to prove that $L$ is homeomorphic to $\mathbb{R}$, all I need to do is show a continuous function $f:L\longrightarrow \mathbb{R}$ such that $f$ is invertible and $f^{-1}$ is also continuous, right?

I am new at this, so I am a bit confused on the Hausdorff part. A topological space is not Hausdorff if there is a pair of distinct points $x,\,y$ such that there are open sets $U$ and $V$ so that $x\in U$ and $y\in V$, but $U\cap V \neq \emptyset$, right?

If what I stated above is true, then I need to find two open sets, one containing the point $(0,0)$ and the other containing $(0,1)$, such that their intersection is not empty? Will that be sufficient?

Thx for the help and sorry for my english.

 

[AKG]

In order to prove that $L$ is homeomorphic to $\mathbb{R}$, all I need to do is show a continuous function $f:L\longrightarrow \mathbb{R}$ such that $f$ is invertible and $f^{-1}$ is also continuous, right?

You don't need to prove L homeomorphic to R, indeed you can't. You need to prove that it is locally homeomorphic to R. What is the definition of "locally homeomorphic"?

I am new at this, so I am a bit confused on the Hausdorff part. A topological space is not Hausdorff if there is a pair of distinct points $x,\,y$ such that there are open sets $U$ and $V$ so that $x\in U$ and $y\in V$, but $U\cap V \neq \emptyset$, right?

No, and the underlined part is wrong. You need to find distinct points x and y such that for every pair of open sets U and V containing x and y respectively, U and V have non-empty intersection.

 

[AiRAVATA]

Sorry for the late reply AKG... thanks anyway for the help...

On the Hausdorff part, I think I got it.

Any point in the upper line is related to any point in the lower line, except the origins. So, every open neigborhood of $(0,1)$ will intersect with an open neighborhood containing the point $(0,0)$. Is this correct?

On the locally homeomorphic part... If I topologize my space with a basis of open subsets $V \subset X$ consisting of the following:

If $(0,1) \notin V$, then $V$ is open as a subset of $\mathbb{R}\times\{0\}$.
If $(0,1) \in V$, then there is an open neighborhood $W \subset (\mathbb{R}\times\{0\})$ of $(0,0)$ such that $V=(W-\{(0,0)\})\cup\{(0,1)\}$.

Then, for every point $p\in X$, there is an open neighborhood $N \subseteq V$, such that there exist a homeomorphism $f:N\longrightarrow U \subset \mathbb{R}$.

Such homeomorphism could be the projection function?

What dou you think? Is this correct?

 

[AKG]

What is the topology for L? You can't say a space is Hausdorff unless you know its topology. How do you know that {(0,0)} isn't an open set? It isn't, don't worry, but how do you know?

What is the definition of locally homeomorphic? Prior to this, I myself had never even seen anyone talk about a local homeomorphism. I just looked it up in my textbook, and if you're using the same definition as my book's, then whatever it is you're doing doesn't show local homeomorphism.