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This painful question is inspired by the question "non-Lie subgroups" . Let $f$ be a discontinuous additive map from $\mathbb{R} \to \mathbb{R}$. Is it possible that the graph of $f$, inside $\mathbb{R}^2$ with the usual topology, is connected? Remember that the graph of a discontinuous function can be connected, as in the Toplogist's sine curve.

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.

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I think the answer is, yes, the graph can be connected.

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.

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".

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. (http://en.wikipedia.org/wiki/Cardinality_of_the_continuum)

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.

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!)

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  • $\begingroup$ Wait a sec, something is wrong with this argument. For any point (x,y) in R^2, the set R^2 \setminus {(x,y)} is open, with boundary {(x,y)}. So a graph which meets every boundary would have to be all of R^2. $\endgroup$
    – David E Speyer
    Commented Oct 30, 2009 at 13:18
  • $\begingroup$ Good point! But I think this doesn't kill the argument. We actually only have to worry about hitting the boundaries of open sets whose complement contains an open set. (since we're trying to avoid the case where graph contained in union of two disjoint open sets.) For these sets the construction still goes through, because unless the open set is the union of vertical lines the projection of the boundary contains an interval. $\endgroup$
    – Martin M. W.
    Commented Oct 30, 2009 at 13:48
  • $\begingroup$ "[U]nless the open set is the union of vertical lines[,] the projection of the boundary contains an interval." Is this clear? That's exactly the sort of issue that made my head hurt when I was thinking about this. $\endgroup$
    – David E Speyer
    Commented Oct 30, 2009 at 15:34
  • $\begingroup$ Yeah, my temples are throbbing now, too! I think the key is we only have to deal with "candidate" open sets where the projection of the boundary onto the x-axis contains an interval. That's because if the graph G = A union B for open sets A,B, then the projections proj(A) and proj(B) onto the x-axis are both open, and must intersect. The set proj(A) intersect proj(B) contains an interval, and for any x in this interval, there is a y such that (x,y) is in boundary(A). I've edited the answer to take this point into account. $\endgroup$
    – Martin M. W.
    Commented Oct 30, 2009 at 17:53
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The construction in the following uses a non-continuous additive map R->R whose graph is connected to build the subgroup described in its title.

Ryuji Maehara, On a connected dense proper subgroup of R^2 whose complement is dense. Proc. AMS, Vol 97, no. 3, July 1986

For further reading, here's another interesting take. The construction is of a connected subgroup of a Lie group that is not path connected. Math Reviews says that there is an error in it, but I think the theorem still holds for Abelian groups.

E.S. Thomas, Connected Subgroups of Lie Groups, Illinois Journal of Mathematics Vol 31, Number 4, Winter 1987

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