# Proof about a Topological space being arc connect

While reading a book i found a topological space described as:

Let $(X,\tau)$ be the topological space formed by adding to the ordinary closed unit interval $[0,1]$ another right end point,say $1*$, with the sets $(a,1)\cup${1*} as a local neighborhood basis.

Then it says that such topological space is arc connected. I found almost exactly the same question here which has yet to be solve,however i'll provide some details.

The book itself states that since [0,1] and [0,1)$\cup${1} are homeomorphic as subspaces,and the subspace topology on [0,1] is Euclidean,X is the union of two compact subspaces and thus compact,by the same reasoning it is arc connected.*

How can such argument prove me that there is a injective path from $1$ to $1*$? Is it possible to explicit such path?

Further details: As the book states:

Path and arc connectednesss relate to the existence of certain continuous functions from the unit interval into a topological space.Continuous functions from the unit interval are called paths,if they are one-to-one they are arcs.

• Are you sure it says this space is arc-connected? This seems like it would be intended as an example of a space that is path-connected but not arc-connected. Jul 9, 2016 at 5:52
• Could it be that the authors use the term arc-connected as synonymous to path-connected? Jul 9, 2016 at 5:54
• Unfortunatelly ,I am its on page 92 of Counterexamples in Topology by Lynn Arthur Steen,and the definition assumed there (of arc) is different from path Jul 9, 2016 at 5:54
• What is their definition of an arc? Jul 9, 2016 at 5:55
• It’s a mistake in Steen & Seebach; the conclusion should be that the space is path connected. There’s at least one other that I know of, though it was also a mistake in the paper from which S&S modified it. Jul 9, 2016 at 5:59

This space is not arc-connected. Indeed, suppose $f:[0,1]\to X$ is an arc such that $f(0)=1$ and $f(1)=1^*$. Then $f(1/2)\in [0,1)$, and $f|_{[0,1/2]}$ is a (reparametrized) path from $1$ to $f(1/2)$ in $[0,1]$. Thus $f([0,1/2])$ must contain all of $[f(1/2),1)$. But by a similar argument, $f([1/2,1])$ also must contain all of $[f(1/2),1)$. This contradicts injectivity of $f$.
• Would you please elaborate why $([1/2,1])$ contains $[f(1/2), 1)$? Jul 29, 2016 at 18:15
• $f|_{[1/2,1]}$ is a path from $f(1/2)$ to $1^*$ in $[0,1)\cup\{1^*\}\cong [0,1]$. Jul 29, 2016 at 18:17