Topological space which is not locally connected In class we defined a locally connected space as a space that has a basis consisting of connected sets. I don't quite understand what a space which is not locally connected would look like. At least one basis element in every basis would have to be disconnected, so the union of two non-empty non-intersecting open sets. That would give us another basis, which would have to contain a disconnected element, and so on. I can't think of a single example of a space that satisfies this. Have I misunderstood the definition? Is there a good, simple example of a space that is locally disconnected? 
I have tried to understand the infinite broom as an example of a connected space that is not locally connected. I can see why it is connected, but I don't understand why it is not locally connected. Could someone help me understand why?
 A: An example of such a space is $\Bbb Q$ with the relative topology. (I think) A general way to get such spaces easily is to consider a connected space and then consider a nice choice of dense subset and look at the relative topology. You could replace the rationals with the irrationals and it would also work.
I think another - more exotic - example is an irrational rotation set. Suppose $\theta\in(0,2\pi)$ is irrational, then let $S = \{e^{in\theta}:n\in\Bbb Z\}$. $S$ is dense in the unit circle but is not locally connected.
A: For concreteness, let's take the comb space $X \subset \mathbb{R}^2$ to be the union
$$([0, 1] \times \{0\}) \cup \left\{\left\{\frac{1}{n}\right\} \times [0, 1] : n \in \mathbb{Z}_+\right\} \cup (\{0\} \times [0, 1])$$
endowed with the subspace topology. (This is similar to the infinite broom, but it is slightly easier to write some things down explicitly.) Certainly the space is connected (even path-connected): Any two points on $X$ are connected by a path that goes down the bristle it is on (as applicable), then along the base, then (again, as applicable) up the bristle the other point is on.
By definition of the topology on $\mathbb{R}^2$, any $\epsilon$-ball $$B := B_{\epsilon}(0, 1) \cap X$$ is open in $X$. Henceforth, assume $\epsilon < 1$, so that $(0, 0) \not\in B$. Now, for any $n > \frac{1}{\epsilon}$, $\left(\frac{1}{\epsilon}, 1\right) \in B$, and so by construction $$B \cap \left\{(x, y) : x < \frac{1}{n + \frac{1}{2}} \right\}$$ and $$B \cap \left\{(x, y) : x > \frac{1}{n + \frac{1}{2}} \right\}$$
comprise a separation of $B$ (in particular, $(0, 1)$ is in the first set, and $\left(\frac{1}{\epsilon}, 1\right)$ is in the second. So, all sufficiently small $\epsilon$-balls $B_{\epsilon}(0, 1) \cap X$ in $X$ are not connected, and since these balls form a neighborhood basis of $X$ at $(0, 1)$, $X$ is not locally connected.
