Open Subset of Locally Path-Connected Space is Locally Path-Connected

Theorem

Let $T = \struct {S, \tau}$ be a locally path-connected topological space.

Let $U \subset S$ be open in $T$.

Then $U$ is locally path-connected in $T$.

Proof

Let $\tau_U$ denotes the subspace topology on $U$ induced by $\tau$.

That is, $\tau_U = \set {\OO \cap U: \OO \in \tau}$.

To show that $\struct {U, \tau_U}$ is a locally path-connected topological space, we must prove that each point $x\in U$ has a local basis consisting of path-connected elements of $\tau_U$.

Let $x$ be an element of $U$.

Since $T$ is locally path-connected, there exists a local basis of $x$ in $T$ consisting of path-connected sets.

Let $\widetilde \BB = \set {\OO_\alpha: \alpha \in \II}$ be such a local basis, where $\II$ is an indexing set.

That is, for each $\alpha \in \II$:

• $\OO_\alpha$ is path-connected

• $\OO_\alpha\in \tau$, and

• $\OO_\alpha$ contains $x$.

Let $\BB = \set {\OO_\alpha \cap U: \alpha \in \II}$.

We claim that $\BB$ gives a local basis for $x$ considered as an element of the topological space $\struct {U, \tau_U}$ (though $\BB$ need not consist of path-connected sets.)

To see this, note that for each $\alpha \in \II$, $\OO_\alpha \cap U$ is open in $U$ by definition of the subspace topology on $U$.

Note also that since $x \in U$ and $x \in \OO_\alpha$:

$x \in U \cap \OO_\alpha$,

so $\BB$ is a family of open sets containing $x$.

To show $\BB$ is a local basis, it remains to prove that any element of $\tau_U$ containing $x$ also contains an element of $\BB$.

Let $A \in \tau_U$ be such that $x \in A$.

Since each element of $\tau_U$ is of the form $\OO \cap U$ where $\OO \in \tau$, there exists $V \in \tau$ such that:

$A = V \cap U$

Since $\tau$ is a topology and $V$ and $U$ are both in $\tau$:

$A \in \tau$

as well.

From the fact that $\widetilde \BB$ is a local basis for $x$ in $\struct {T, \tau}$, we may therefore find an element $\alpha \in \II$ such that

$\OO_\alpha \subseteq A$

By Set Intersection Preserves Subsets/Corollary:

$\OO_\alpha \cap U \subseteq A \cap U$

From Intersection with Subset is Subset, since $A \subseteq U$:

$A \cap U = A$

Therefore

$\OO_\alpha \cap U \subseteq A$

as required.

We now use the local basis $\BB$ for $x$ to produce a local basis for $x$ consisting of path-connected sets.

Let $A \in \BB$.

Since $A$ is a neighborhood of $x$ and $\widetilde B$ is a local basis for $x$ considered as an element of the topological space $\struct {T, \tau}$, there exists an element $\alpha \in \II$ such that $\OO_\alpha \in \widetilde B$ satisfies

$\OO_\alpha \subseteq A$.

Using the Axiom of Choice, we may therefore define a function $\phi : \BB \to \II$ such that

$\OO_{\map \phi A} \subseteq A$.

We claim that $\map \phi \BB$ gives a local basis for $x$ in $U$ consisting of path-connected sets.

First, note that for all $A \in \BB$,

$\map \phi A \in \tau$

and

$\map \phi A \subseteq A \subseteq U$.

It follows from the definition of the subspace topology $\tau_U$ that

$\map \phi A = \map \phi A \cap U \in \tau_U$

Second, note that since $\map \phi A$ is path-connected in $T$ and $\map \phi A \subseteq U$, it is immediate from the definition of path-connectedness that $\map \phi A$ is also path-connected when considered as a subset of the topological space $\struct {U, \tau_U}$.

Finally, note that for any element $V\in \tau_U$ such that $x \in V$, since $\BB$ is a local basis for $x$ considered as an element of $\struct {U, \tau_U}$, there exists $A \in \BB$ such that

$A \subseteq V$

Since $\map \phi A \subseteq A$, it follows that

$\map \phi A \subseteq V$

$\map \phi \BB$ is therefore a local basis for $x$ in the topological space $\struct {U, \tau_U}$ consisting of path-connected sets as required.

$\blacksquare$

Axiom of Choice

This theorem depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.

Also see

• Open Subset of Locally Connected Space is Locally Connected, an analogous result for connected components