# 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 $\sigma:[0,1]\to X$.

## 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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