Would it be true if I said that the image of x is contained in f(X)?

[amanda_ou812]

Homework Statement

Let (X,d) and (Y, p) be metric spaces and f : X -> Y a function. Prove that f is continuous at p-0 if and only if for every ε > 0, there exists a δ > 0 so that the image of Bd(p-0; δ) is contained in Bp( f (p-0);ε).

Homework Equations

The Attempt at a Solution

For the forward implication, I feel like I am almost there.

The statement f is continuous at p-0 implies for every e>0 there is a delta>0 so that when d(p, p-0)< delta this implies p( f(p), f(p-0))<e. But d(p, p-0)< delta implies x is contained in B (sub d) (p-0; delta) and p( f(p), f(p-0))<e implies f(p) is contained in B (sub p) (f(p-0); e).

Here is where I would like to say that notice that the image of x is contained in f(x). Then I can conclude that the image of B (sub d) is contained in B (sub p)

Then, I was going to reverse it for the other implication.

 

[HallsofIvy]

For any set X, the "image of x" is defined as the set of all y such that y= f(x) for all x in X. Normally, we use the notation f(X) to mean the "image of x". So the image of x is not just "contained in" fix), it is f(x).