Fix . We calculate
using the limit laws, which proves is continuous at .
Since was arbitrary is continuous on .
- Could be the image of a path.
- Could be the image of a path.
- Could not be the image of a path as it is not contiguous.
- Could be the image of a path.
Take
which is continuous and satisfies has and .
This is a domain as it is open and path connected.
This is a domain as it is open and path connected.
This is not a domain as it is not path connected.

This is a domain as it is open and path connected.
- Fix . Since is open we can find such that . Since is open we can find such that . Put . Then . Since was arbitrary, the intersection is open.
The red and blue sets above are both domains. However, their intersection is not path connected. The intersection of two domains need not be a domain.