## Proposition 6.4.1: Continuity and Topology

*f*be a function with domain

*in*

**D***. Then the following statements are equivalent:*

**R***f*is continuous- If
is open, then the inverse image of every open set under**D***f*is again open. - If
is open, then the inverse image of every open interval under**D***f*is again open. - If
is closed, then the inverse image of every closed set under**D***f*is again closed. - If
is closed, then the inverse image of every closed interval under**D***f*is again closed. - The inverse image of every open set under
*f*is the intersection ofwith an open set.**D** - The inverse image of every closed set under
*f*is the intersection ofwith a closed set.**D**

### 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 f^{-1}(**U)**
**D** is
again open. Take any x_{0}
f^{-1}(**U)**.
That is equivalent to saying that f( x_{0})
**U**.
Since **U** is open, we can find an >
0 such that the -
neighborhood of f( x_{0})
is contained in **U**. For this fixed we
can use the continuity of f to pick a >
0 such that

- if | x - x
_{0}| < then | f(x) - f( x_{0}) | <

This implies that the -
neighborhood of x_{0}is
contained in f^{-1}(**U)**.
Hence, the inverse image of the arbitrary set **U **is open.

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

- if | x - x
_{0}| < then this set is contained in the set f^{-1}( f( x_{0}) - , f( x_{0}) + )

or in other words

- if | x - x
_{0}| < then | f(x) - f( x_{0}) | <

**(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

- f
^{-1}( comp(**U**) ) = comp( f^{-1}(**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.