Proposition: Continuity and Topology

• Let f be a function with domain D in R. Then the following statements are equivalent:
  1. f is continuous
  2. If D is open, then the inverse image of every open set under f is again open.
  3. If D is open, then the inverse image of every open interval under f is again open.
  4. If D is closed, then the inverse image of every closed set under f is again closed.
  5. If D is closed, then the inverse image of every closed interval under f is again closed.
  6. The inverse image of every open set under f is the intersection of D with an open set.
  7. The inverse image of every closed set under f is the intersection of D with a closed set.

Proof:

(1) => (2): Assume that f is continuous on an open set D. Let U be an open set in the range of f. We need to show that (U) D is again open. Take any (U). That is equivalent to saying that f( ) U. Since U is open, we can find an > 0 such that the - neighborhood of f( ) is contained in U. For this fixed we can use the continuity of f to pick a > 0 such that

• if | x - | < then | f(x) - f( ) | <

This implies that the - neighborhood of is contained in (U). Hence, the inverse image of the arbitrary set U is open.

(2) => (1): Assume that the inverse image (U) of every open set U is open. Take any point U and pick an > 0. Then the - neighborhood of f( ) is an open set, so that it's inverse image is again open. That inverse image contains , and since it is open it contains a - neighborhood of for some > 0. But that is exactly what we want:

• if | x - | < then this set is contained in the set ( f( ) - , f( ) + )

or in other words

• if | x - | < then | f(x) - f( ) | <

(2) <=> (3): This follows immediately from the fact that every open set in the real line can be written as the countable union of open intervals.

(2) <=> (4): This follows immediately by looking at complements, i.e. from the fact that

• ( comp(U) ) = comp( (U) )

That equality should be proved as an exercise.

(4) <=> (5): This follows again by combining the two previous remarks.

(6), (7): This proof is very similar to the proof of (2) <=> (1). In fact, where in that previous proof have we used the fact that the domain D of the function is open ? The details are left as an exercise again.