A counterexample in topology Semi-local simple connectedness is a property that arises in Algebraic Topology in the study of covering spaces, namely, it is a necessary condition for the existence of the universal cover of a topological space X. It means that every point $x \in X$ has a neighborhood $N$ such that every loop in $N$ is nullhomotopic in $X$ (not necessarily through a homotopy of loops in $N$). The way I see it, the prefix "semi-" is refers more to "simply connected" than to "locally", since if such $N$ exists, all other neighborhoods of $x$ inside $N$ also have the property, so each point has a fundamental system of (open) neighborhoods for which the property holds (EDIT see Qiaochu's comment). Instead it isn't true that a semi-local simply connected space is locally simply connected (i.e each point has a fundamental system of open, simply connected neighborhoods): take the space $$ X = \frac{H \times I}{ \sim } $$ where $H$ is the "Hawaiian earring", or infinite earring (which is an example of non semi-locally simply connected space) and $\sim$ is the equivalence that identifies $H \times \{0\}$ to one point.
However I was interested in finding another type of counterexample. Consider the topological property (call it $*$) consisting in the existence, for all $x \in X$, of a simply connected, not necessarily open, neighborhood of $x$. We have $$ \text{semi-local simple connectedness} \Leftarrow * \Leftarrow \text{local simple connectedness} \wedge \text{simple connectedness} $$ I am wondering if $$ \text{semi-local simple connectedness} \Rightarrow * $$ holds. Intuitively it shouldn't, but I'm having trouble finding a counterexample. For example, the space $X$ described above won't work because it is simply connected (even contractible). It seems to me that, if a counterexample does exist, it must have local pathologies (to ensure that a certain point $x$ doesn't have a simply connected neighborhood), and globally the space should allow loops close to $x$ to be nullhomotopic, but in such a way that every neighborhood $N$ of $x$ contains a small enough loop that will not contract in $N$. EDIT Also, I am looking for a counterexample which is a locally path connected space (a previous answer showed a counterexample without this property.. but @answerer it was interesting anyway, you shouldn't have deleted it!)
But perhaps I'm wrong, and the two assertions are equivalent, or maybe I am missing something very simple. If anybody has any ideas please share, thank you!
 A: Call the "interesting" point in the Hawaiian earring $q$ (the point where all the circles intersect).  In your space $X$, glue the quotiented point $\{H \times \{0\}\}$ to the point $(q,1) \in H \times I$, essentially bending your cone around in a loop.  Call the resulting quotient space $Y$, and the glued point $p$.  Note that $p$ could be described either as $(q,1)$ or as $\{H \times \{0\}\}$.  ($Y$ can also be described as the Mapping Torus of the map that takes the entire Hawaiian earring to the point $q$).
$Y$ is semi-locally simply connected but doesn't satisfy property $∗$.
To see $Y$ is semi-locally simply connected, note that any point has a neighborhood that is disjoint from a set of the form $H \times \{x\}$ for some $x \in I$.  The inclusion of such a neighborhood into $Y$ is nullhomotopic (simply retract the portion contained in $H \times [0,x)$ to $p$, then follow the strong deformation retraction of $X$ to $\{H \times \{0\}\}$), which certainly implies that any loop contained in it is nullhomotopic in $Y$.
But $p$ itself has no simply-connected neighborhood.  Any such neighborhood $N$ contains a loop $L \times \{1\}$ in $H \times \{1\}$, which could only be contracted in $N$ by including all of $L \times I$ in $N$.  In particular, you'd have to include all of $\{q\} \times I$ in $N$,  But $\{q\} \times I$ is itself a loop which isn't nullhomotopic in $Y$, and so also isn't nullhomotopic in $N$.  Thus, $N$ cannot be simply connected.
