Question

Let ε>0 and δ>0 and a Є R. Show that Vε(a)nVδ(a) and Vε(a)UVδ(a) are γ neighborhoods of a for appropriate values of γ.

Answer 1

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$\varepsilon\geq 0$ and $\delta\geq 0$ and $a\in\mathbb{R}$. Show that $V_{\varepsilon}(a)\cap V_{\delta}(a)$ and $V_{\varepsilon}(a)\cup V_{\delta}(a)$ are $\gamma$ neighborhoods of $a$ for appropriate values of $\gamma$.

Solution. Recall that by definition of topology in $\mathbb{R}$ a neighbourhood $V_{t}(a)$ is the interval $(a-t,a+t)$ for each $t>0$, so it consists of all $x\in\mathbb{R}$ such that

$a-t<x<a+t$

which is the same as

$-t<x-a<t.$

In particular,

$V_{\varepsilon}(a)=(a-\varepsilon,a+\varepsilon)\qquad V_{\delta}(a)=(a-\delta,a+\delta).$

1) First we will show that

$V_{\varepsilon}(a)\cap V_{\delta}(a)=V_{\min\{\varepsilon,\delta\}}(a).$

Indeed, the intersection

$V_{\varepsilon}(a)\cap V_{\delta}(a)$

consists of all $x\in\mathbb{R}$ for which both of the following pairs of inequalities hold:

$-\varepsilon<x-a<\varepsilon\qquad\text{and}\qquad-\delta<x-a<\delta.$

Notice that

$-\varepsilon<x-a\qquad\text{and}\qquad-\delta<x-a$

if anf only if

$-\min\{\varepsilon,\delta\}<x-a.$

Similarly,

$x-a<\varepsilon\qquad\text{and}\qquad x-a<\delta$

if anf only if

$x-a<\min\{\varepsilon,\delta\}.$

In other words,

$V_{\varepsilon}(a)\cap V_{\delta}(a)$

consists of all $x\in\mathbb{R}$ for which

$-\min\{\varepsilon,\delta\}<x-a<\min\{\varepsilon,\delta\},$

and so

$V_{\varepsilon}(a)\cap V_{\delta}(a)=V_{\min\{\varepsilon,\delta\}}(a).$

2) Now we wil prove that

$V_{\varepsilon}(a)\cup V_{\delta}(a)=V_{\max\{\varepsilon,\delta\}}(a).$

Indeed, the union

$V_{\varepsilon}(a)\cup V_{\delta}(a)$

consists of all $x\in\mathbb{R}$ for which at least one of following pairs of inequalities hold:

$-\varepsilon<x-a<\varepsilon\qquad\text{and}\qquad-\delta<x-a<\delta.$

Notice that at least one the following inequalities hold

$-\varepsilon<x-a\qquad\text{or}\qquad-\delta<x-a$

if anf only if

$-\max\{\varepsilon,\delta\}<x-a.$

Similarly, at least one the following inequalities hold

$x-a<\varepsilon\qquad\text{or}\qquad x-a<\delta$

if anf only if

$x-a<\max\{\varepsilon,\delta\}.$

In other words,

$V_{\varepsilon}(a)\cap V_{\delta}(a)$

consists of all $x\in\mathbb{R}$ for which

$-\max\{\varepsilon,\delta\}<x-a<\max\{\varepsilon,\delta\},$

and so

$V_{\varepsilon}(a)\cap V_{\delta}(a)=V_{\max\{\varepsilon,\delta\}}(a).$

Answer.

$V_{\varepsilon}(a)\cap V_{\delta}(a)=V_{\min\{\varepsilon,\delta\}}(a),\qquad V_{\varepsilon}(a)\cap V_{\delta}(a)=V_{\max\{\varepsilon,\delta\}}(a).$