Show that the inclusion map i:Q -> R defined by i(q)=q for all q in Q, is continuous where both Q (rational numbers) and R(real numbers) are given the order topology.© BrainMass Inc. brainmass.com December 24, 2021, 7:48 pm ad1c9bdddf
SOLUTION This solution is FREE courtesy of BrainMass!
Please See Attached
is the set of all real numbers. For any , we define and . As an order topology, and are base open sets and any open set in is the finite intersections of the base open sets.
To show that inclusion map is continuous, we need to prove that and are open in . From the definition, we know is an identity map for any .
We consider the open set in . We have two cases.
Case 1: , then is an open set in as an ordered topology.
Case 2: , then since is dense in , we can find a decreasing number series , such that and . Now we define the set . Since , then is an open set in . Next I claim that . For each , since , then . On the other side, for any , . Since , we can find some , such that . Then . Therefore, . So is a union of open sets in and hence is also an open set in .
From the above two cases, is an open set in . Similarly, is also an open set in .
Since is an identity map, then for any two base open sets , we have
and it is also open.
Therefore, for any open set in , is open in .
Hence is continuous with respect to order topology.