Answer:
Proved
Step-by-step explanation:
To prove that every point in the open interval (0,1) is an interior point of S
This we can prove by contradiction method.
Let, if possible c be a point in the interval which is not an interior point.
Then c has a neighbourhood which contains atleast one point not in (0,1)
Let d be the point which is in neighbourhood of c but not in S(0,1)
Then the points between c and d would be either in (0,1) or not in (0,1)
If out of all points say d1,d2..... we find that dn is a point which is in (0,1) and dn+1 is not in (0,1) however large n is.
Then we find that dn is a boundary point of S
But since S is an open interval there is no boundary point hence we get a contradiction. Our assumption was wrong.
Every point of S=(0, 1) is an interior point of S.
The sum of -5a and 3 is greater than 1
Answer:
important, valuable, nice,
I belive the unit rate per goal is 5.7
Answer:
1,800
Step-by-step explanation:
