In general closure of a path connected set is not path connected. For example, the graph of sin(1/x) on (0,1] .
However, if we require the original set to be open, does >that make the clousre path connected? So my conjecture is: > >"If E is an open, path connected set, then closure(E) is path connected."
Intereting discussion about limit of graph connectivity problem - Following .
Check these similar questions:
https://math.stackexchange.com/questions/286896/is-the-closure-of-path-connected-set-path-connected
https://math.stackexchange.com/questions/162867/when-is-the-closure-of-a-path-connected-set-also-path-connected
Ok, and what about this answer from Topology Atlas (http://at.yorku.ca/topology/):
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2006&task=show_msg&msg=0668.0001
Intereting discussion about limit of graph connectivity problem - Following .
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