A topological space \(X\) is locally connected at \(x\in X\) if for every nbhd \(U\) of \(x\) there is a connected nbhd \(V\) of \(x\) with \(V\sub U\). We say that \(X\) is locally connected if it is locally connected at each \(x\in X\).
A trivial space is locally connected. A discrete space is locally connected. An Euclidean space \(\br ^{n}\) is locally connected. The topologist’s sine curve is not locally connected.
A topological space \(X\) is locally connected if and only if the components of every open subset of \(X\) are open in \(X\).
Proof. Assume that \(X\) is locally connected and \(U\) is open in \(X\). If \(C\) is a component of \(U\) and \(x\in C\), then \(U\) is a nbhd of \(x\) in \(X\) so there is a connected nbhd \(V\) of \(x\) in \(X\) with \(x\in V\sub U\). Then \(V\sub C\) so \(C\) is a nbhd of \(x\) in \(X\). Since \(C\) is a nbhd of each \(x\in C\), it follows that \(C\) is open in \(X\).
Now assume that the components of every open subset of \(X\) are open in \(X\). Let \(x\in X\) and \(U\) be a nbhd of \(x\). Let \(U’\) be an open nbhd of \(x\) with \(U’\sub U\) and \(V\) be the component of \(U’\) containing \(x\). Then \(V\) is a connected nbhd of \(x\) with \(V\sub U\). Thus \(X\) is locally connected. □
Every open subspace of a locally connected topological space is locally connected.
Let \(f:X\to Y\) be a continuous closed surjection. If \(X\) is locally connected, then \(Y\) is locally connected.
Proof. Let \(U\) be open in \(Y\) and \(C\) be a component of \(U\). We will show that \(f^{-1}\bof C\) is open in \(X\). Since \(f\) is a closed surjection, we have
\[ f\bof {X\sem f^{-1}\bof C}=Y\sem C, \]
so it will follow then that \(Y\sem C\) is closed in \(Y\) and consequently that is \(C\) is open.
Let \(x\in f^{-1}\bof C\). Then \(x\in f^{-1}\bof U\), which is open in \(X\). Let \(V\) be the component of \(f^{-1}\bof U\) that contains \(x\). Then \(f\of x\in f\bof V\) and \(f\bof V\) is connected so \(f\bof V\sub C\) and hence \(V\sub f^{-1}\bof C\). Since \(V\) is open, \(f^{-1}\bof C\) is a nbhd of \(x\). Since \(f^{-1}\bof C\) is a nbhd of each \(x\in f^{-1}\bof C\), it follows that \(f^{-1}\bof C\) is open, as required. □
The above result also holds when \(f\) is a continuous open surjection.
Let \(X:=\set 0\cup \bn \) with discrete topology and
\[ Y:=\set 0\cup \set {\frac {1}{n}:n\in \bn } \]
with the topology inherited from \(\br \). If \(f:X\to Y\) is defined by \(f\of 0:=0\) and \(f\of n:=1/n\) for each \(n\in \bn \), then \(f\) is a continuous surjection. The space \(X\) is locally connected, but \(Y\) is not.
Let \(X_{\ga }\) be a topological space for each \(\ga \in A\) and \(X:=\prod _{\ga \in A}X_{\ga }\). Then \(X\) is locally connected if and only if \(X_{\ga }\) is locally connected for each \(\ga \in A\) and all but finitely many \(X_{\ga }\) are connected.
Proof. Assume that \(X\) is locally connected. Let \(\gb \in A\) and \(x\in X_{\gb }\). We show that \(X_{\gb }\) is locally connected at \(x\). Let \(U_{\gb }\) be an open nbhd of \(x\) in \(X_{\gb }\). Then \(p_{\gb }^{-1}\bof {U_{\gb }}\) is open in \(X\). Let \(\xi =\cur {\xi _{\ga }}_{\ga }\in p_{\gb }^{-1}\bof {U_{\gb }}\) with \(\xi _{\ga }=x\). There is a connected nbhd \(V\) of \(\xi \) in \(X\) with \(V\sub p_{\gb }^{-1}\bof {U_{\gb }}\). Then
\[ x\in p_{\gb }\bof V\sub U_{\gb } \]
and \(p_{\gb }\bof V\) is a connected nbhd of \(x\) since \(p_{\gb }\) is continuous and open. Thus \(X_{\gb }\) is locally connected at \(x\).
Now we show that all but finitely many of \(X_{\ga }\) are connected. Let \(C\) be a component of \(X\). Then \(C\) is open so there is finite \(A’\sub A\) and open \(U_{\ga }\in X_{\ga }\) for every \(\ga \in A\) such that
\[ \emp \neq B:=\bigcap _{\ga \in A’}p_{\ga }^{-1}\bof {U_{\ga }}\sub C. \]
If \(\ga \in A\sem A’\), then \(p_{\ga }\bof B=X_{\ga }\) so \(p_{\ga }\bof C=X_{\ga }\), which implies that \(X_{\ga }\) is connected.
Now assume that \(X_{\ga }\) is locally connected for each \(\ga \in A\) and \(A’\sub A\) is finite and such that \(X_{\ga }\) is connected for every \(\ga \in A\sem A’\). Let \(\xi =\cur {\xi _{\ga }}_{\ga \in A}\in X\). To show that \(X\) is locally connected at \(\xi \), it suffices to show that for every finite \(B\sub A\) and an open nbhd \(U_{\ga }\) of \(\xi _{\ga }\) for every \(\ga \in B\), the set
\[ V:=\bigcap _{\ga \in B}p_{\ga }^{-1}\bof {U_{\ga }} \]
contains a connected nbhd of \(\xi \). Given a set \(V\) as described above, let \(V_{\ga }\) be a connected nbhd of \(\xi \) with \(V_{\ga }\sub U_{\ga }\) for every \(\ga \in B\) and let \(V_{\ga }\) be any connected nbhd of \(\xi _{\ga }\) for each \(\ga \in A’\sem B\). Let
\[ C:=\bigcap _{\ga \in A’\cup B}p_{\ga }^{-1}\bof {V_{\ga }}=\prod _{\ga \in A}V_{\ga }, \]
where \(V_{\ga }=X_{\ga }\) for \(\ga \in A\sem \cur {A’\cup B}\). Then \(C\) is a nbhd of \(\xi \) contained in \(V\) and \(C\) is connected by Theorem 3.1.12, since \(V_{\ga }\) is connected for each \(\ga \in A\). □
A topological space \(X\) is locally path-connected at \(x\in X\) if for each nbhd \(U\) of \(x\) there exists a path-connected nbhd \(V\) of \(x\) with \(V\sub U\). We say that \(X\) is locally path-connected if it is locally path-connected at each \(x\in X\).
A topological space \(X\) is locally path-connected if and only if the path-components of every open subspace of \(X\) are open.
Proof. Assume that the path-components of every open subspace of \(X\) are open. If \(x\in X\) and \(U\) is an open nbhd of \(x\) in \(X\), then the component of \(U\) that contains \(x\) is the required path-connected nbhd of \(x\) that is contained in \(U\). Thus \(X\) is locally path-connected.
Assume that \(X\) is locally path-connected. Let \(U\) be open in \(X\) and \(P\) be a path-component of \(U\). If \(x\in P\), there is a path-connected nbhd \(V\) of \(x\) with \(V\sub U\). Then \(P\cup V\) is path-connected so \(P\cup V=P\) and so \(P\) is a nbhd of \(x\). Since \(P\) is a nbhd of each \(x\in P\), it is open. □
A topological space \(X\) is locally path-connected if and only if it has a basis consisting of path-connected sets.
Let \(X\) be a locally-path connected space. Then each path-component of \(X\) is clopen and is a component of \(X\).
Proof. Let \(P\) be a path component of \(X\). Then \(P\) is open. Since each other path component is also open, it follows that \(P\) is closed. Since \(P\) is clopen, no proper superset of \(P\) can be connected. Since \(P\) is connected, it is a component. □
The topologist’s sine curve is connected, but it is not locally path-connected. It’s path components are neither closed nor open.
A connected locally-path connected space is path-connected.
Let \(f:X\to Y\) be a continuous surjection that is open or closed. If \(X\) is locally path-connected, then \(Y\) is also locally path-connected.
Let \(X_{\ga }\) be a topological space for each \(\ga \in A\) and \(X:=\prod _{\ga \in A}X_{\ga }\). Then \(X\) is locally path-connected if and only if \(X_{\ga }\) is locally path-connected for each \(\ga \in A\) and all but finitely many \(X_{\ga }\) are path-connected.