Topology Q: Show f is Continuous in X with d and A

[mbarby]

hi all,
i am studying from croom's introduction to topology book. i came across such a question. and i don't have a clue as to how to start .
Let X be a metric space with metric d and A a non-empty subset of X. define f:X->IR by :

f(x): d(x,A), x E X (x is an element of X)
show that f is continuous.

also if you can point out a solution book for this book that would be rather nice, considering i am computer scientist studying the topic at home..
thx.

 

[micromass]

Hi mbarby!

You'll need to prove that

$$d(x,y)<\delta~\Rightarrow~|d(x,A)-d(y,A)|<\varepsilon$$

Can you first prove that

$$-d(x,y)<d(x,A)-d(y,A)<d(x,y)$$

 

[mbarby]

to prove that i use the triangular inequality
d(x,A) <= d(x,y)+d(y,A)
d(x,A) - d(y,A) <= d(x,y)

--->
-d(x,y) <= d(x,A) - d(y,A) <= d(x,y)

but honestly i couldn't connect it to any kind of a proof :/ ...

 

[micromass]

Doesnt that immediately imply

$$|d(x,A)-d(y,A)|\leq d(x,y)$$

and this would imply continuity...

 

[blue_raver22]

I would suggest approaching this problem by first reviewing the definition of continuity in topology. In general, a function f is continuous at a point a if for any open set U containing f(a), there exists an open set V containing a such that f(V) is a subset of U. This definition can also be extended to a function being continuous on a set A if it is continuous at every point in A.

In this specific problem, we are given a function f(x) = d(x,A), where d is the metric on X and A is a non-empty subset of X. To show that f is continuous, we need to show that it is continuous at every point in X. Let a be any arbitrary point in X and let U be an open set containing f(a). We want to find an open set V containing a such that f(V) is a subset of U.

Since A is a non-empty subset of X, there exists at least one point in A, say b, such that d(a,b) < d(a,A). This follows from the definition of distance between a point and a set. Now, consider the open ball B(b, r) centered at b with radius r = d(a,b) - d(a,A). By construction, this open ball contains a and is contained in A. Therefore, for any x in B(b,r), we have d(x,A) = d(x,b) < d(a,b) - d(a,A) + d(a,A) = d(a,b). This implies that f(x) < d(a,b) for all x in B(b,r).

Now, if we take V to be the open ball B(a,r) centered at a with radius r = d(a,b) - d(a,A), we have V contained in B(b,r) and therefore, f(V) is a subset of B(b,r), which is a subset of U. Hence, we have shown that for any point a in X, we can find an open set V containing a such that f(V) is a subset of U. This proves that f is continuous on X.

As for a solution book, I would recommend looking for solutions to specific problems online or consulting with a topology professor or fellow students for guidance. Good luck with your studies!