# In topology,Find all points of the set B={(-2,-2),(1,1),(0,-1),(0,3)}, which are contained in the closed ball B((0,0),2)

Find all points of the set B={(-2,-2),(1,1),(0,-1),(0,3)}, which are contained in the closed ball B((0,0),2) in the metric space (R^2,d1),where d1 is defined by the formula

d1((x1,y1),(x2,y2))=|x1-x2|+|y1-y2|.

Does anybody clearly explain and solve by formally ?

Many thanks!

It's pretty straight forward isn't it? You are told that the distance between points (a, b) and (c, d) is |a- c|+ |b- d| so the distance from (0, 0) to (x, y) is |x- 0|+ |y- 0|= |x|- |y|. What is the distance from (0, 0) to (-2, -2)? From (0, 0) to (1, 1)? From (0, 0) to (0, -1)? From (0, 0) to (0, 3)?

• Could you draw and more then detailed to solving, please ? Jan 10, 2019 at 16:35
• Could you more than explain ? because I couldnt be understand.....Could you solving by formally by step by step ? Many thanks... Jan 11, 2019 at 15:50

By definition you can check that only $$(0,-1)$$ has distance less than $$2$$ from the origin.

• Could you draw and more then detailed to solving, please ? Jan 9, 2019 at 18:41
• Could you more than detail and solving by formally ? Many thanks... Jan 11, 2019 at 15:49

A required formal and detailed step-by-step solution is still short, simple and straightforward.

$$d_1((0,0),(-2,2))=|0-(-2)|+|0-2|=2+2=4>2$$, so $$(-2,2)\not\in B((0,0),2)$$.

$$d_1((0,0),(1,1))=|0-1|+|0-1|=1+1=2$$, so $$(1,1)\in B((0,0),2)$$.

$$d_1((0,0),(0,-1))=|0-0|+|0-(-1)|=0+1=1<2$$, so $$(0,-1)\in B((0,0),2)$$.

$$d_1((0,0),(0,3))=|0-0|+|0-3|=0+3=3>2$$, so $$(0,3)\not\in B((0,0),2)$$.