Potential connected non-Lie subgroup - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T21:34:06Z http://mathoverflow.net/feeds/question/3289 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3289/potential-connected-non-lie-subgroup Potential connected non-Lie subgroup David Speyer 2009-10-29T17:09:47Z 2009-10-30T20:17:47Z <p>This painful question is inspired by the question "<a href="http://mathoverflow.net/questions/3157/non-lie-subgroups" rel="nofollow">non-Lie subgroups</a>" . Let R denote the real numbers. Let f be an discontinuous additive map from R --> R. Is it possible that the graph of f, inside R^2 with the usual topology, is connected? Remember that the graph of discontinuous function can be connected, as in the <a href="http://en.wikipedia.org/wiki/Topologist%27s%5Fsine%5Fcurve" rel="nofollow">Toplogist's sine curve</a>.)</p> <p>I was hoping to show that any connected subgroup of a Lie group is a Lie group, thus answering the previous question, and I couldn't get rid of this horrible example. </p> http://mathoverflow.net/questions/3289/potential-connected-non-lie-subgroup/3335#3335 Answer by Martin M. W. for Potential connected non-Lie subgroup Martin M. W. 2009-10-29T20:54:19Z 2009-10-30T20:17:47Z <p>I think the answer is, yes, the graph can be connected.</p> <p>By definition, if the graph G is not connected, then we can find disjoint nonempty open sets A and B, such that G is contained in A union B. In particular, this implies no point in G can be contained in the boundary of A. So if we can construct an additive function f whose graph intersects the boundary of any potential separating open set A, we'll have shown the graph is connected.</p> <p>Before constructing this function, note a technical point. Not all open sets are candidates for separating G. If G = A union B for nonempty open sets A,B, then the projections proj(A) and proj(B) onto the x-axis are both open, and must intersect. In turn this implies the projection of the boundary of A contains an interval. Call open sets with this property "candidate sets". </p> <p>To make a function f whose graph intersects the boundary of all candidate sets, consider a basis H for R as a vector space over Q. This set has cardinality of the reals. Now note that the set of all open sets in R^2 also has cardinality of the reals. (<a href="http://en.wikipedia.org/wiki/Cardinality_of_the_continuum" rel="nofollow">http://en.wikipedia.org/wiki/Cardinality_of_the_continuum</a>)</p> <p>Put these two sets (basis H, all open sets) in 1-1 correspondence, so for each h in H, we have an open set O(h). If O(h) is not a "candidate set," let f(h)=0. Otherwise, using the fact that O(h) is a candidate set, we can always find a nonzero rational q, and a real y such that (qh,y) is in the boundary of O(h). Define f(qh)=y. Doing this for all elements of H will determine a unique additive function f on the reals.</p> <p>The graph of f, by construction, is connected since it intersects the boundary of every candidate separating open set in R^2. (And it's not continuous--if it were, it would miss the boundaries of a lot of open sets!)</p>