## Proposition 6.4.4: Images of Compact and Connected sets

*f*is a continuous function on a domain

*, then:*

**D**- the image of every compact set is again compact.
- the image of every connected set is again connected.

### Proof:

**1.** Note that the image of a closed set is not necessarily
closed for continuous functions, and the image of a bounded set
is not necessarily bounded. However, the image of a close *and
*bounded set is again closed and bounded (under continuous
functions). Despite this, the proof is fairly easy: Recall that
a set **D** is compact if every open cover of **D** can
be reduced to a finite subcover.

Let **A** be an open cover of the set *f( D)*. Since

*f*is continuous, the collection

- {f
^{-1}(**U**):**U****A**}

is a collection of open sets that cover **D**. Since **D**
is compact, this collection can be reduced to a finite subcover,
say:

- f
^{-1}( U_{1}), f^{-1}( U_{2}), ..., f^{-1}( U_{n}),

But then the sets U_{1}, U_{2}, ..., U_{n} cover
the set *f( D)*. Hence, every open cover of

*f(*can be reduced to a finite subcover. Therefore,

**D**)*f(*is compact.

**D**)
**Note:** We have used the fact that the inverse image of open
sets is open. That, however, is only true if the original domain
**D** of the function is also open. Can you modify this proof
so that it still works for a not necessarily open domain **D
**?

**2.** This prove is again simple. The idea is as follows:
suppose **U** **D**
is connected, but *f( U)* is not connected. Then

*f(*with**U**) = A B*A*,*B*open and disjoint

Since *f* is continuous, f^{-1}(A)
and f^{-1}(B) are
both open. They are clearly disjoint, and their union makes up
all of **U**. But then **U** is not connected, which is
a contradiction. Thus, the image of every connected set under
a continuous function is connected.

**Note: **Just as before, this proof is not completely correct.
It does reflect the major idea, but some technicalities are missing.
Why is this proof technically not correct ? Can you fix it ?