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)# ?

Question

1294 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-13T01:31:14

#gamma=min{epsilon, delta}#

If #epsilon < delta #, then #V_(epsilon)(a) cap V_(delta) (a)=V_(epsilon)(a)#, so #gamma=epsilon#.

If #epsilon=delta#, then #V_(epsilon)(a) cap V_(delta) (a)=V_(epsilon)(a)=V_(delta)(a)#, so #gamma=epsilon=delta#.

If #epsilon > delta#, then #V_(epsilon)(a) cap V_(delta) (a)=V_(delta)(a)#, so #gamma=delta#.

Hence, #gamma=min{epsilon, delta}#

I hope that this was clear.

Answer 2 · 2017-04-13T10:52:36

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

If #a < b# then

#V_(epsilon)(a)cap V_(delta)(b)=(b-delta,a+epsilon)=V_(gamma)(c)#,

where

#gamma=((a+epsilon)-(b-delta))/2# and #c=((a+epsilon)+(b-delta))/2#.

If #b < a# then

#V_(epsilon)(a)cap V_(delta)(b)=(a-epsilon,b+delta)=V_(gamma)(c)#,

where

#gamma=((b+delta)-(a-epsilon))/2# and #c=((b+delta)+(a-epsilon))/2#.

I hope that this was clear.