Thread: Local path-connectedness definition

A topological space X is said to be locally path-connected (or locally arcwise connected) if, for any point x, every open neighborhood U of x contains an OPEN path-connected set V containing x. This is essentially the definition given in Massey, A Basic Course in Algebraic Topology, p. 36, and other books use an equivalent definition (for example, this is exactly Corollary 1.19, p. 28, in Rotman, An Introduction to Algebraic Topology).

The definition in Massey actually says, " . . . if each point has a basic family of arcwise connected neighborhoods," but I assume the word "neighborhood" here means an OPEN set containing the point, as it usually does, and the above expresses this, I believe.

Later on, however, in Exercise 2.1 on p. 123, Massey asks the reader to prove that the definition above is equivalent to the definition WITH THE WORD "OPEN" OMITTED. That is, the condition is that U above must contain a path-connected set V containing x (NOT nec. an open one). This makes no sense to me, because such a set could then be {x} itself, and thus any space would be locally path connected.

I am obviously missing something here. Can anyone enlighten me?

Thanks.

Originally Posted by J Graham

A topological space X is said to be locally path-connected (or locally arcwise connected) if, for any point x, every open neighborhood U of x contains an OPEN path-connected set V containing x. This is essentially the definition given in Massey, A Basic Course in Algebraic Topology, p. 36, and other books use an equivalent definition (for example, this is exactly Corollary 1.19, p. 28, in Rotman, An Introduction to Algebraic Topology).

The definition in Massey actually says, " . . . if each point has a basic family of arcwise connected neighborhoods," but I assume the word "neighborhood" here means an OPEN set containing the point, as it usually does, and the above expresses this, I believe.

Later on, however, in Exercise 2.1 on p. 123, Massey asks the reader to prove that the definition above is equivalent to the definition WITH THE WORD "OPEN" OMITTED. That is, the condition is that U above must contain a path-connected set V containing x (NOT nec. an open one). This makes no sense to me, because such a set could then be {x} itself, and thus any space would be locally path connected.

I am obviously missing something here. Can anyone enlighten me?

Thanks.

In topology, a neighborhood is basically defined in two different ways.

1. "U is a neighborhood of x" refers to "U is an open set containing x" (Munkres's Topology, etc)
2. "A is a neighborhood x" refers to "A contains an open set containing x" (wiki)

For 2, an open set in A containing x is an open subset of "int A (interior of A)", which is the largest open set contained in a neighborhood A.

Definition. A space X is locally path connected at a point p in X if every open set containing p contains a path connected open set containing p. The space X is locally path connected provided that it is locally path connected at each of its points.

The below lemma is basically equivalent to what Massey's book ask you to prove.

Lemma. A space X is locally path connected if and only if for each open subset O of X, each path component of O is an open set.

Proof.

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Assume X is locally path connected. Let O be an open set in X; let P be a path component of O; let x be a point of P. By the definition of a locally path connected, we can choose a path connected open set U of X such that . Since U is path connected, it must lie entirely in the path component P of O. Thus, P is open in X.

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Assume a path component of an open set O is open. Given a point x of X and an open set O of x, let P be the path component of O containing x. Since P is an open set containing x and is contained in O, it satisfies the conditions for a locally path connected for X. Thus, X is locally path connected.

Note.
Not all path components are open. For instance, the path components of as a subspace of a standard topology on are not open. A topologist's sine curve is another example that the path components are not necessarily open.

Thanks for the clarification, Aliceinwonderland.

The lemma you prove is part of the same exercise (2.1, p. 123) in Massey that I cited, and I had no trouble with that because, unfortunately, it didn't really relate to my confusion.

What did resolve the situation was your second definition of "neighborhood," which makes it clear that the existence of a path-connected neighborhood of x, in the sense of a (not necessarily open) set that contains an open subset containing x, obviously implies the existence of an open path-connected set containing x.

Hence, this second definition shows that the requirement for local path-connectedness that the path-connected set containing x be open is still satisfied using this definition.

Thanks again. I knew it was my mistake and not Massey's. The book is excellent -- I highly recommend it -- and I've found only a few very minor misprints so far, nothing of any consequence.