# Let a is real and epsilon>0. Given V_epsilon(a)={x : |x-a|<epsilon}. How to find gamma>0 so that V_gamma(a)=V_epsilon(a)nnV_delta(a) ?

Apr 13, 2017

$\gamma = \min \left\{\epsilon , \delta\right\}$

#### Explanation:

If $\epsilon < \delta$, then ${V}_{\epsilon} \left(a\right) \cap {V}_{\delta} \left(a\right) = {V}_{\epsilon} \left(a\right)$, so $\gamma = \epsilon$.

If $\epsilon = \delta$, then ${V}_{\epsilon} \left(a\right) \cap {V}_{\delta} \left(a\right) = {V}_{\epsilon} \left(a\right) = {V}_{\delta} \left(a\right)$, so $\gamma = \epsilon = \delta$.

If $\epsilon > \delta$, then ${V}_{\epsilon} \left(a\right) \cap {V}_{\delta} \left(a\right) = {V}_{\delta} \left(a\right)$, so $\gamma = \delta$.

Hence, $\gamma = \min \left\{\epsilon , \delta\right\}$

I hope that this was clear.

Apr 13, 2017

Please see the explanation below for my response to dill's comment.

#### Explanation:

If $a < b$ then

${V}_{\epsilon} \left(a\right) \cap {V}_{\delta} \left(b\right) = \left(b - \delta , a + \epsilon\right) = {V}_{\gamma} \left(c\right)$,

where

$\gamma = \frac{\left(a + \epsilon\right) - \left(b - \delta\right)}{2}$ and $c = \frac{\left(a + \epsilon\right) + \left(b - \delta\right)}{2}$.

If $b < a$ then

${V}_{\epsilon} \left(a\right) \cap {V}_{\delta} \left(b\right) = \left(a - \epsilon , b + \delta\right) = {V}_{\gamma} \left(c\right)$,

where

$\gamma = \frac{\left(b + \delta\right) - \left(a - \epsilon\right)}{2}$ and $c = \frac{\left(b + \delta\right) + \left(a - \epsilon\right)}{2}$.

I hope that this was clear.