|Jan30-13, 01:07 PM||#1|
Topology of the unit interval
I need help with something basic but I'm not sure how to handle it. The doubt is about how to consider the topology of the unit interval I=[0,1] inherited of the real line with its usual topology (intervals of the type (a,b)).
I think that is just to pay attention to the definition, I mean, the open subsets of 'I' would be the intersection of a usual open interval and 'I'. In this way, 'I' itself would be a open subset of the inherited topology, and all the sets of the form [0,x), (a,b) and (y,1] -with 0 < x,a,b,y <1 - would be open sets of the inherited topology.
Please, can anyone tell me if I'm right?
Thanks in advance.
|Jan30-13, 02:06 PM||#2|
Sounds about right.
Note that some sets in the subspace are open sets even if they aren't open in the larger space.
For instance, [0, 1] is closed in R. But when we consider [0, 1] as a subspace, it's open (because the entire topological space is required to be open in any topology).
Similarly, [0, 1), which is neither open nor closed in R is open in [0, 1].
|Jan30-13, 02:17 PM||#3|
Thank you Tac-Tics
|Similar Threads for: Topology of the unit interval|
|Basic topology proof of closed interval in R||Calculus & Beyond Homework||1|
|Unit tangent, unit normal, unit binormal, curvature||Calculus & Beyond Homework||4|
|Should I do a unit in topology?||Academic Guidance||2|
|The usual topology is the smallest topology containing the upper and lower topology||Topology and Analysis||2|
|The usual topology is the smallest topology containing the upper and lower topology||Calculus & Beyond Homework||0|