Struggling with a Metric Space Problem: Show A is Closed

[mansi]

Here’s a problem I’ve been struggling with, for a while….
If (X,d) is a metric space and f:X-->X is a continuous function, then show that A={ x in X : f(x)=x} is a closed set.
One possible way that I can think of is defining a new function g(x) = f(x)-x .Then A={x in X : g(x) =0}. Now {0} is closed, so by property of continuity, inverse image of a closed set is closed. Thus A is closed.
However, I’d like to do this problem using the basic definition of closed sets, i.e. show that the complement of A is open….so if I pick up an element ‘p’ not belonging to A, I must find a radius r so that B(p,r) is contained in the complement of A. This is where I’m getting stuck… please help !

 

[AKG]

Use the definition of continuity. If f is continuous at p, then there is an open ball of radius r around p such that for all x in that ball, f(x) is arbitrarily close to f(p). Make that arbitrary distance, $\epsilon$ < |f(p)|.

 

[wais]

First of all, don't get discouraged if you're struggling with this problem. Metric spaces can be difficult to work with, but with practice and persistence, you will be able to master them.

Now, to show that A is closed using the definition of closed sets, we need to show that the complement of A, denoted by A', is open. In other words, for every point p in A', there exists a radius r such that the open ball B(p,r) is contained in A'.

Let p be any point in A'. This means that p is not in A, so f(p) is not equal to p. Since f is continuous, we know that for any epsilon greater than 0, there exists a delta greater than 0 such that if d(x,p) < delta, then d(f(x),f(p)) < epsilon.

Now, since f(p) is not equal to p, we can choose epsilon to be half the distance between f(p) and p, i.e. epsilon = d(f(p),p)/2. Then, there exists a delta > 0 such that if d(x,p) < delta, then d(f(x),f(p)) < d(f(p),p)/2.

Let r = delta. Then, for any x in B(p,r), we have that d(x,p) < r = delta. This means that d(f(x),f(p)) < d(f(p),p)/2. By the triangle inequality, we have that d(x,p) <= d(x,f(x)) + d(f(x),p). Combining this with the previous inequality, we get that d(x,p) < d(x,f(x)) + d(f(x),p) < d(f(p),p)/2 + d(f(p),p)/2 = d(f(p),p).

This means that d(x,f(p)) < d(f(p),p), so x is not in A. Therefore, x is in A'. Since x was arbitrary, this holds for all points in B(p,r). Hence, B(p,r) is contained in A'.

Therefore, we have shown that for any point p in A', there exists a radius r such that B(p,r) is contained in A'. This means that A' is open, and hence A is closed.

I hope this helps! Keep practicing and don't hesitate to ask for help if you get stuck on any