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.