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Show that $\mathbb{R}$ with standard topology is Hausdorff.

For any $x,y\in\mathbb{R}$ it is possible to define $\mathscr{U}_x=(x-\epsilon,x+\epsilon)$ for $\epsilon>0$ and $\mathscr{U}_y=(y-\delta,y+\delta)$ so that $\mathscr{U}_x\cap\mathscr{U}_y=\emptyset$.

Question:

Is this proof right? If not. How should I answer the question? What tools should I use?

Thanks in advance!

Pedro Gomes
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    To make the proof right, verify that the intersection is empty. For that you will have to show how to choose $\delta$ and $\epsilon$. For example, $\delta = \epsilon = |x-y|/2$. – GEdgar Oct 19 '18 at 16:34
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    "...it is possible..." This assertion is not a proof on its own, but needs a proof. – drhab Oct 19 '18 at 16:43

3 Answers3

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The argument is not complete, until you find $\varepsilon$ and $\delta$ such that $$ (x-\varepsilon,x+\varepsilon)\cap(y-\delta,y+\delta)=\emptyset $$ Since $x\ne y$, it is not restrictive to assume $x<y$. In order the intersection above is empty, it's sufficient that $$ x+\varepsilon<y-\delta $$ that is, $\varepsilon+\delta<y-x$. Take $$ \varepsilon=\delta=\frac{y-x}{3} $$ and you're done.

egreg
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Assume $x <y$. Then $$x<\frac{x+y}{2}<y$$

Here the sets $(-\infty,\frac{x+y}{2})$ and $(\frac{x+y}{2},\infty)$ makes the separation for $x$ and $y$

Chinnapparaj R
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An easier way to do it is to choose both $$\epsilon = \delta = \frac{x+y}{2} > 0$$

Then define,

$$U_x = (x - \epsilon, x + \epsilon) \subseteq \mathcal R$$ $$U_y = (y - \epsilon, y + \epsilon) \subseteq \mathcal R$$

As you can see, both are open intervales; therefore, they are open sets in the standard (usual) topology in $R$. In other words,

$$U_x, U_y \in \mathcal T$$

Furthermore,

$$U_x \: \cap \: U_y = \mathcal \phi$$

Ahmad Haitham Ibrahim
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