Is it possible to construct a function f:R→R such that f(x+y)=f(x)+f(y) and f is not continuous?
Answer
"Fix a basis for R as a Q-vector space" (which exists under the axiom of choice, but under weaker axioms I have no idea what happens). The condition f(x+y)=f(x)+f(y) is equivalent to f being Q-linear, so you're asking if there exists a non-trivial discontinuous map of Q-vector spaces between R and itself. If you map the basis elements to other basis elements in a discontinuous way, you will obtain such a map.
Added : A quick way to see that "a discontinuous way of doing it" exists, is that the set of Q-linear maps that are also R-linear has cardinality of the reals, where as the set of all Q-linear maps (or in other words, the number of maps between the basis elements) has cardinality |R||R|. To understand why, well of course the basis for R as a Q-vector space has cardinality ≤|R|, but if it was countable, then R would be countable because all of its elements would be a linear combination of elements in a countable set with countably many possible coefficients. Since the basis for R as a Q-vector space is uncountable, the set of all maps from a basis to another also is. Therefore there must be at least one map between the two bases which generates a discontinuous linear map.
Hope that helps,
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