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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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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, that this implies no point in G can be contained in the boundary of A. So to make the graph connected, it's enough to if we can construct an additive function f whose graph intersects the boundary of everyany potential separating open set (other than R^2 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 \emptyset). 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 do somake 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 the union of vertical linesnot a "candidate set," let f(h)=0. Otherwise, using the graph of any f will intersect its boundary by virtue of being fact that O(h) is a graph, so candidate setf(h) to anything. Otherwise, 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 open set in R^2. (And it's not continuous--if it were, it would miss the boundaries of a lot of open sets!)

4 explicitly dealt with open sets that are unions of vertical lines.

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 open sets A and B, such that G is contained in A union B. In particular, that implies no point in G can be contained in the boundary of A.

So to make the graph connected, it's enough to construct an additive function f whose graph intersects the boundary of every open set (other than R^2 and \emptyset). To do so, 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). We If O(h) is the union of vertical lines, the graph of any f will intersect its boundary by virtue of being a graph, so set f(h) to anything. Otherwise, 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 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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