Proving Interior Point of A if x in A: R^p

[onie mti]

i was given that A= B(0,1)={x in R^2| 2-norm of x is less than 1}
show that
x in R^p is an interior point of A iff x in A

my work

i managed to prove this one way and struggling with the other, well a bit confused
i said assume that x is an interior pt of A
then:
B(x,r) is a subset of A.
but x is in B(x,r) HENCE x is in A.

now suppose the converse, that is x in A
then the 2-norm of x is less than 1
now I am stuck:(

 

[jvicens]

To prove the converse, we need to show that x is an interior point of A. This means that there exists an open ball centered at x that is completely contained within A.

Since x is in A, we know that the 2-norm of x is less than 1. Let's call this 2-norm value r. This means that B(x,r) is a ball centered at x with radius r.

We need to show that B(x,r) is a subset of A. Let y be any point in B(x,r). This means that the distance between x and y is less than r. Since r is the 2-norm of x, this also means that the 2-norm of y is less than r.

But we know that the 2-norm of y is less than 1, since it is less than r and r is the 2-norm of x. This means that y is also in A. Since y was an arbitrary point in B(x,r), this means that B(x,r) is a subset of A.

Therefore, x is an interior point of A, as desired.