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A topological space $X$ is called locally path connected if for every $x \in X$ and open set $U$ containing $x$, there is a path connected open set $V \subseteq U$ such that $x \in V$.

Is every contractible metric space locally path connected ?

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No. For instance, let $K\subset\mathbb{R}$ be the Cantor set and let $X=\{(tx,t):x\in K, t\in[0,1]\}\subset\mathbb{R}^2$. Then $X$ is contractible (by the map $H:X\times[0,1]\to X$ given by $H(y,t,s)=(sy,st)$) but is not locally connected at any point other than $(0,0)$.

More generally, given any space $A$, you can consider the cone $X=A\times[0,1]/A\times\{0\}$ on $A$, which is always contractible but contains $A\times(0,1]$ as an open subspace and so will not be locally (path-)connected unless $A$ is. In general $X$ may not be metrizable, but it will be if $A$ is a compact metric space.

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Let $A = \{a\in\mathbb{Q}\mid -1 \le a \le 1\}$.

Let $X=\{(x,y)\in \mathbb{R}^2\mid y=ax\;\text{for some}\;a\in A\;\text{and}\;0\le x \le 1\}$.

Let the metric on $X$ be inherited from the Euclidean metric on $\mathbb{R}^2$.

Then $X$ is contractible (to the origin), but not locally path connected at any point of $X$ other than the origin.

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Another example is the comb $\left([0,1] \times \{0\}\right) \cup \left( \bigcup_{n=1}^{\infty}\{1/n\} \times [0,1]\right)\cup \left(\{0\} \times [0,1]\right) \subset \mathbb{R}^2$. It is not locally path-connected on points in the leftmost side of the comb (except on the base point, where it is), but it is clearly homotopically equivalent to $[0,1]$, and thus to a point.

If instead of taking the comb roots to be $\{0\} \cup\{1/n\}$ we take the rational numbers, we get yet another example. This time, it is not locally path-connected in any point other than the base points.

If we make this last comb a triangular comb and concatenate them like this:

enter image description here

we get a contractible metric space which is not locally path-connected anywhere.

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