locally path connectedness

Question

While studying covering spaces , hatcher mentioned the "shrinking wedge of circles" this space is locally path connected as I was told , but I wasn't able to prove it nor to see it, it looks like comb space to me which is not locally path connected , can anyone help me prove it locally path connected ? I appreciate your help

Answer 1

In the shrinking wedge of circles, the only problematic point for the local connectedness is the point $p$ where all the circles meet (for the other points are locally as an interval). But in $p$ we have also a base of path-connected neighbourhoods, namely the intersection with any ball centered at $p$:

And this is enough for the local path-connectedness.

Answer 2

Let $X$ be the shrinking wedge of circles and let $P\in X$ be the wedge point. For any $Q\in X$ with $Q\ne P$ there is an open neighborhood $U_Q$ containing $Q$ such that $U_Q\cap X$ is homeomorphic to the open interval $(0,1)$ which is path connected.

Now, try to show that if $B$ is any open ball centered at $P$, and $Q\in B\cap X$, then there is a path from $Q$ to $P$ which lies in $B\cap X$. There are essentially two cases here: either $Q$ lies on a circle which is completely contained in $B$, or $Q$ lies on a circle which is not.

Answer 3

The shrinking wedge is a union of circles, all through the origin. The only problem to local path connectedness can be at the origin. But a typical nbhd of that point is a union of circle arcs trough the origin, which is arc connected (any point there is connected to the origin by the arc it is contained in).