Spooky geometry III

Construct a path-connected metric space X and a discontinuous function f:X\to \Bbb{R} such that f\circ \sigma is continuous for any continuous path \sigma:[0,1]\to X.

Spooky geometry III

3 thoughts on “Spooky geometry III

  1. If X is locally path-connected and first-countable, it is true that a function f:X\to Y such that its composition with any path is continuous is itself continuous. We will show a counterexample where X is locally path-connected but not first-countable.

    Let X be the subspace of \ell^\infty of points of the form \lambda e_n for some \lambda\in[0,1] and natural n; that is, X is the union of the unit segments corresponding to the “coordinate axes”.

    Let f:\ell^\infty\to\Bbb{R} be any linear functional such that e_n\mapsto n and consider its restriction to X. Obviously this restriction is discontinuous at 0 since f is unbounded. However, the image of any path \sigma:[0,1]\to X is contained, by compactness, in a finite-dimensional subspace of the form \langle e_1,\dots,e_k\rangle, so the restriction of f to this subspace is bounded, and therefore continuous.

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    1. The above solution is wrong, since the example is both locally path-connected and first-countable (any metric space is first-countable!). In fact, it is not true that the image of a continuous path \sigma is contained in a finite-dimensional subspace. For instance, consider a path \sigma which goes from 0 to e_1/2 and returns in [0,1/2], then from 0 to e_2/4 and returns in [1/2, 3/4], and so on. This is continuous in [0,1) by the pasting lemma, and if we set \sigma(1)=0, the fact that e_n/2^{n+1}\longrightarrow 0 implies \sigma is continuous at 1, too. What led to this mistake was the fact that the image of a compact set in a CW-complex intersects at most finitely many cells, but the obvious CW structure on the set X does not induce the same topology we are considering: it is not even metrizable, since the complex is not locally finite!

      Here is a correct solution, due to Pablo, which is a variation on the comb space. Let X=([-1,1]\times \{0\})\cup(([-1,1]\cap\Bbb{R}\setminus\Bbb{Q}) \times [0,1]) and define f(x,y)=0 if x<0 and f(x,y)=y if x\geq 0. This is obviously discontinuous at any point (0,y) with positive y, but f\circ\sigma is continuous for any path \sigma. Notice that this space is not locally path-connected.


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