Proving f Continuous for Topological Problem on A, B Open/Closed

[qinglong.1397]

Homework Statement

Suppose X = A\cupB where A and B are closed sets. Suppose f : (X, T_X) \rightarrow (Y, T_Y ) is a map such that f|_A and
f|_B are continuous (where A and B have their subspace topologies). Show that f is continuous. What happens if A and B
are open? What happens if A or B is neither open nor closed?


T_X means the topology on set X; T_Y the topology on Y. f|_A means the restriction of f on A; f|_B the restriction on B.

Homework Equations

The Attempt at a Solution

I do not know how to prove, so is there anyone who can give me the answer? Thank you very much!

 

[Dick]

Nobody is going to 'give you the answer'. But just think about real numbers. Suppose A=[0,1) and B=[1,2]. So X=[0,2]. Pick f(x)=1 for x in A and f(x)=2 for x in B. So f is continuous on both A and B. But it's not continuous on X. Why wouldn't this happen if A and B were closed?

 

[qinglong.1397]

Nobody is going to 'give you the answer'. But just think about real numbers. Suppose A=[0,1) and B=[1,2]. So X=[0,2]. Pick f(x)=1 for x in A and f(x)=2 for x in B. So f is continuous on both A and B. But it's not continuous on X. Why wouldn't this happen if A and B were closed?

Thanks for your hint. But I still do not know how to generalize it. Would you please give me another hint?

 

[Dick]

Thanks for your hint. But I still do not know how to generalize it. Would you please give me another hint?

Not until you figure out what the last hint means. Why isn't the f I gave you continuous on X?

 

[qinglong.1397]

Not until you figure out what the last hint means. Why isn't the f I gave you continuous on X?

OK. The reason is that A=[0, 1) does not interact with B=[1, 2], so the function f can take different values at 1. If A=[0, 1], according to the definition of function, f has only one value at 1, so it is continuous.

Am I right?

 

[Dick]

OK. The reason is that A=[0, 1) does not interact with B=[1, 2], so the function f can take different values at 1. If A=[0, 1], according to the definition of function, f has only one value at 1, so it is continuous.

Am I right?

No. f isn't continuous. I think you mean to say A and B don't 'intersect' not 'interact', but that still doesn't make it continuous. And f does have only one value at x=1. f(1)=2 because x=1 is in B. The problem is that x=1 is the limit of a sequence of points in x_n in A, so the value of f(x_n)=1, but f(1)=2. Look at the definition of continuity in terms of sequences again. Can you give an example of such a sequence? And why couldn't this have happened if A were closed?