# Topology - Is every non-empty open subspace of a locally path connected space also locally path connected?

Let $$X$$ be a non-empty locally path connected topological space and $$A$$ a non-empty open subspace of $$X$$. Is $$A$$ also locally path connected? I remember seeing this in a lecture but it was used implicitly and without justification. I cannot find a counterexample. So I'm trying to prove it.

Consider any point $$x\in A$$. To show that $$A$$ is locally path connected, it suffices to find a neighborhood base at $$x$$ consisting of path connected subsets of $$A$$. Since $$X$$ is locally path connected, we can find a neighborhood base $$\mathscr{B}$$ at $$x$$ which consists of path connected subsets of $$X$$. Let $$\mathscr{B}'=\{B\in \mathscr{B}:B\subseteq A\}$$. Then $$\mathscr{B}'$$ is a neighborhood base at $$x$$ in $$A$$. Indeed, since $$A$$ is open in $$X$$, it is a neighborhood of $$x$$ in $$X$$, hence contains some member $$B_1$$ of $$\mathscr{B}$$, meaning that $$\mathscr{B}'\neq \emptyset$$.

Then consider any neighborhood $$U$$ of $$x$$ in $$A$$. We know that $$U=A\cap V$$ for some neighborhood $$V$$ of $$x$$ in $$X$$. There exists $$B_2\in \mathscr{B}$$ with $$B_2\subseteq V$$. By definition of neighborhood base, there exists $$B_3\in \mathscr{B}$$ with $$B_3\subseteq B_1\cap B_2$$. It follows that $$B_3\subseteq A\cap V=U$$. Hence every neighborhood of $$x$$ in $$A$$ contains some member of $$\mathscr{B}'$$, proving the claim.

Now consider any $$B\in \mathscr{B}'$$ and any two points $$y,z\in B$$. Since $$B$$ is a path connected subset of $$X$$, there exists a continuous function $$f:[0,1]\to B$$ with $$f(0)=y,f(1)=z$$. But $$f([0,1])\subseteq B\subseteq A$$, meaning that $$f$$ is also a $$yz$$-path in $$A$$. We conclude that $$A$$ is locally path connected.

Is my proof correct? Thanks for any comment.

I might consider breaking the paragraph at “Then consider” to help readability.

It's a combination of two simple observations.

• Being path-connected is intrinsic, not dependent on the subspace.
• If each point has a local base of sets with some property $$P$$, and $$O$$ is open, we can assume that for each $$x \in O$$, all the elements of the local base are a subset of $$O$$ too.

This will also do for local connectedness and local compactness (in its strong form) etc.