----- > [!proposition] Proposition. ([[the ordered square is connected but not path-connected]]) > The [[ordered square]] $S \subset \mathbb{R}^{2}$ is [[connected]] but not [[path-connected]]. > [!proof]- Proof. ([[the ordered square is connected but not path-connected]]) > Being a [[linear continuum]], it is clear that the [[ordered square]] $I \times I$ is [[connected]] ([[linear continuums in the order topology and the intervals they contain are connected|see here]]). > But it is not [[path-connected]]: let $p=(0,0)$ and $q=(1,1)$. We suppose there is a [[parameterized curve]] $f:[0,1] \to S$ joining $p$ and $q$ and derive a contradiction. $f([0,1])$ must contain every point of $S$, by the [[intermediate value theorem]]. Therefore, for each $x \in I$, the set $U_{x}:=f^{-1}\big( \{ x \} \times (0,1) \big)$ is a nonempty subset of $[0,1]$; by [[continuity]] it is open in $[0,1]$. Choose, for each $x \in I$, a [[rational]] number $q_{x}$ belonging to $U_{x}$. Since the sets $U_{x}$ are disjoint, the map $x \mapsto q_{x}$ is an [[injection]] from $I$ to $\mathbb{Q}$. This contradicts the fact that the interval $I$ is uncountable. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```