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.
== Proof ==
=== 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
Category:Open Subset of Locally Connected Space is Locally Connected
Category:Topology
Category:Locally Connected Spaces