Let $X$ be a topological space, $Y \subseteq X$. If $X$ is a metric space, then the closure of $\bar{Y}$ is the set of limit points of sequences in $Y$. I thought of the following generalisation (now for $X$ arbitrary again). Define the path-closure of $Y$ in $X$ to be:
$$Y^P = \bigcup_S \gamma([0,1]) = \bigcup_S \gamma(1)$$
where $S$ = $\{ \gamma:[0,1] \to X : \gamma([0,1)) \subseteq Y \}$. Thus, $Y^P$ is the set of limit points of "paths in $Y$". It can be shown that $Y \subseteq Y^P \subseteq \bar{Y}$.
This notion of closure is much more restrictive, so naturally, it is also much weaker. It turns out that path-closure is not idempotent i.e. we do not necessarily have $(Y^P)^P = Y^P$.
Here is an example. Let $X = \mathbb{R}^2$. I will use some unusual notation: if $x \in \mathbb{R}, I \subseteq \mathbb{R}$, let $(x,I)$ be the obvious interval in $\mathbb{R}^2$. (Similarly for $(I,y)$.) Consider:
$$Y = \Bigg(\bigcup_{x\in\mathbb{Q}\cap[0,1)} \Big(x,[-1,0)\Big) \Bigg) \cup \Bigg(\bigcup_{x\in\mathbb{R}\setminus\mathbb{Q}\cap[0,1)} \Big(x,(0,1]\Big) \Bigg)$$
(Double-sided comb with no shaft and dense/co-dense set of teeth on either side.)
Then,
$$Y^P = \Bigg(\bigcup_{x\in\mathbb{Q}\cap[0,1)} \Big(x,[-1,0]\Big) \Bigg) \cup \Bigg(\bigcup_{x\in\mathbb{R}\setminus\mathbb{Q}\cap[0,1)} \Big(x,[0,1]\Big) \Bigg)$$
but:
$$(Y^P)^P = \Bigg(\bigcup_{x\in\mathbb{Q}\cap[0,1]} \Big(x,[-1,0]\Big) \Bigg) \cup \Bigg(\bigcup_{x\in\mathbb{R}\setminus\mathbb{Q}\cap[0,1]} \Big(x,[0,1]\Big) \Bigg)$$
Note that $(1,0) \in (Y^P)^P \setminus Y^P$. Hence, this concept does not currently satisfy what we would expect from a notion of closure.
Let $Y^{P^2}$ denote $(Y^P)^P$. We will also write $Y^{P^n}$ for $n$ arbitrary.
Question: Let the full path-closure of a subspace $Y$ be $\bigcup_n Y^{P^n}$. Is this operation idempotent on $\mathcal{P}(X)$?
Perhaps this is a solution, but the definition of $Y^{P^n}$ is quite unpleasant to deal with.
Question: Is there a better definition of the notion of path-closure? Is this just completely useless?