- Define what it means for a topological space \(X\) to be:
Connected
Locally connected
- Give, with proof, an example of a space that is connected but not locally connected.
Solution
1. A topological space \(X\) is connected if it cannot be written as\[ X=U\cup V \]with \(U,V\subseteq X\) disjoint, nonempty, and open.
This is the definition of connectedness. Equivalently, the only subsets of \(X\) that are both open and closed are \(\varnothing\) and \(X\).
2. A topological space \(X\) is locally connected if for every \(x\in X\) and every open neighborhood \(U\) of \(x\), there is a connected open set \(V\) such that\[ x\in V\subseteq U. \]
This is the neighborhood-basis definition of local connectedness: every point has a basis of connected open neighborhoods.
3. Consider the comb space\[ C = ([0,1]\times\{0\}) \cup (\{0\}\times[0,1]) \cup \bigcup_{n\ge1}\left(\left\{\frac1n\right\}\times[0,1]\right) \subset\mathbb R^2. \]Then \(C\) is path connected, hence connected.
Let \((x,y)\in C\). If \(y>0\), then the definition of \(C\) forces either \[ x=0 \] or \[ x=\frac1n \] for some \(n\ge1\). In either case the vertical segment from \((x,y)\) to \((x,0)\) lies entirely in \(C\). The horizontal segment from \((x,0)\) to \((0,0)\) lies in \[ [0,1]\times\{0\}\subset C. \] If \(y=0\), only the horizontal segment is needed. Thus every point of \(C\) can be joined by a path to \((0,0)\), so \(C\) is path connected. Every path-connected space is connected.
4. The comb space \(C\) is not locally connected at\[ p=\left(0,\frac12\right). \]
Let \[ U=C\cap B\left(p,\frac14\right). \] This is an open neighborhood of \(p\) in the subspace \(C\). Suppose, contrary to 4, that there is a connected open set \(V\subseteq C\) with \[ p\in V\subseteq U. \]
Because \(V\) is open in \(C\), there is some \(\varepsilon>0\) such that \[ C\cap B(p,\varepsilon)\subseteq V. \] Choose \(n\) sufficiently large that \[ \frac1n<\varepsilon. \] Then \[ q=\left(\frac1n,\frac12\right) \in C\cap B(p,\varepsilon) \subseteq V. \]
Every point of \(U\) has second coordinate strictly between \(1/4\) and \(3/4\). Hence \(U\) does not meet the horizontal base \([0,1]\times\{0\}\). Therefore every point of \(V\subseteq U\) has first coordinate in \[ D=\{0\}\cup\left\{\frac1m:m\ge1\right\}. \] The continuous projection \[ \operatorname{pr}_1:V\to\mathbb R \] has image contained in \(D\) and contains both \[ 0=\operatorname{pr}_1(p) \qquad\text{and}\qquad \frac1n=\operatorname{pr}_1(q). \] If \(V\) were connected, its continuous image \(\operatorname{pr}_1(V)\) would be connected in \(\mathbb R\). Any connected subset of \(\mathbb R\) containing \(0\) and \(1/n\) contains the whole interval \([0,1/n]\). But \(D\) contains no nontrivial interval, a contradiction.
Thus no connected open neighborhood of \(p\) is contained in \(U\). By 2, \(C\) is not locally connected at \(p\).
5. Therefore \(C\) is a connected space that is not locally connected.
Connectedness follows from 3 and failure of local connectedness from 4.