Open Subset of Locally Connected Space is Locally Connected

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

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

Then $U$ is locally connected.

Lemma
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 connected topological space, we must prove that each point $x\in U$ has a local basis consisting of connected elements of $\tau_U$.

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

Since $T$ is locally connected, there exists a local basis of $x$ in $T$ consisting of 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 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 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 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 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 connected in $T$ and $\map \phi A \subseteq U$, $\map \phi A$ is also connected when considered as a subset of the topological space $\struct {U, \tau_U}$ by our lemma above.

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 connected sets as required.

Also see

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