Mathematics Prelims

January 25, 2009

Continuous Functions from the Reals to the Rationals

Filed under: Topology — cjohnson @ 6:45 pm

Using the ideas of connectedness previously discussed, we can show that a continuous function from \mathbb{R} to \mathbb{Q} must be constant.  We will do this by showing that if we have such a continuous function, we’d have a surjection between a connected space and a disconnected space, which we showed in the last post can’t happen.

Suppose f : \mathbb{R} \to \mathbb{Q} is a non-constant map; we’re not making any assumptions about continuity or surjectivity, just that the image of the map has at least two distinct points.  Let q_1, q_2 be two distinct points in the image of \mathbb{R}; q_1, q_2 \in f(\mathbb{R}) : q_1 \neq q_2.  Now note that f is a surjection from the reals to its image, f(\mathbb{R}).  Assume, without loss of generality, q_1 < q_2.  Since these points are distinct, there must be an irrational number in \mathbb{R}, call it r, between them.  Let Q_1 = \{ q \in f(\mathbb{R}) : q < r \} and Q_2 = \{ q \in f(\mathbb{R}) : q > r \}.  Notice that these are open in the subspace topology of f(\mathbb{R}) as \mathbb{Q} \cap (-\infty, r) is open in \mathbb{Q} and Q_1 = f(\mathbb{R}) \cap (\mathbb{Q} \cap (-\infty, r)), and likewise for Q_2.  These two sets are clearly disjoint and cover f(\mathbb{R}) and are non-empty as q_1 \in Q_1, \, q_2 \in Q_2.  This means that f(\mathbb{R}) is a disconnected space.  We know that f is a surjection to f(\mathbb{R}) a disconnected space.  This is impossible if f is continuous, so we have if f : \mathbb{R} \to \mathbb{Q} is non-constant it is not continuous, which by contrapositive means that if f is continuous, it must be constant.

(This is a modified version of the proof in Crossley, with some gaps filled in.)


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