Neighborhoods in real analysis problem

Show that if $a,b \; \epsilon\; \mathbb R$, then there exists $\varepsilon$ neighborhoods U of $a$ and $V$ of $b$ such that $U \cap V = \varnothing $.

I have already defined the sets $V_{\varepsilon}(a):= \{x\epsilon R: |x-a| < \varepsilon\}$ and $U_{\varepsilon}(b):= \{y\epsilon R: |y-b| < \varepsilon\}$ but I don't know how to proceed further. Any help would be appreciated.

Answer

Draw a picture. If $\epsilon=\frac{|b-a|}{3}$, it is clear that the intervals $(a-\epsilon, a+\epsilon)$ and $(b-\epsilon, b+\epsilon)$ have no point in common.

If we want to be very formal, suppose to the contrary that $|b-x|\lt \epsilon$ and $|x-a|\lt \epsilon$. Then by the Triangle Inequality
$$|b-a|\le |b-x|+|x-a|\lt 2\epsilon \lt |b-a|,$$
which is impossible.