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A path in set $E$ is a continuous function $$\gamma:[a,b]\longrightarrow E, $$ with $A:=\gamma(a)$ its starting point and $B:=\gamma(b)$ its ending point.
Any two points $A, B$ in a path-connected set can be connected by a path, with $A, B$ as its starting and ending points.
A Jordan path $\gamma$ is an injective path.
Any two points $P, Q$ in a Jordan-path connected set can be connected by a Jordan path.

Question: For any path-connected set $E$, is $E$ always Jordan-path connected?

When there are at most countably many self-intersecting points, it's possible to 'cut' the loop and make the path a Jordan one. But I got problems when there are uncountably many self-intersecting points on the path.

Eric Monlye
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1 Answers1

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Consider $S$ the Sierpiński space. It has two points $x$ and $y$, where $y$ is open (formally, the open sets are $\varnothing, \{y\}$ and $\{x,y\}$).

The map $\gamma \colon [0,1] \to S$ given by $\gamma(0) = y$ and $\gamma(1) = x$ is continuous: the preimage of the only non-trivial open set $y$ is $[0,1)$, which is open.

This shows that $S$ is path-connected, but it can't be Jordan path connected by cardinality considerations.

This answer by Hagen von Eitzen shows that if $E$ is geometric enough (and by that I mean Hausdorff) then the answer is yes.

qualcuno
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