What are the properties of the set of the Real Numbers without the Integers? This question came up in a lunchtime discussion with coworkers.  None of us are professional mathematicians or teachers of math.  I apologize for any incorrect math or sloppy terminology.
We were discussing getting from one number to another along the real number line.  The challenge is that we removed the integers from the number line.
So can an ant get from 0.5 to 1.5 by crawling along this "broken" line?
Searching the web, we discovered this may be a problem in topology, somehow (possibly) related to something called the "Long Line".  This is far outside of any of our knowledge so we would appreciate an explanation at roughly the basic calculus level.
 A: The space you describe 
$...(-(n+1),-n)...(-2,-1)(-1,0)(0,1)(1,2)...(n,n+1)...$
is homeomorphic to countably infinite ("$\omega$") many copies of $(0,1)$. The Long Line you refer is made of uncountably infinite many copies of $[0,1)$. It can be thought of as a generalization of the nonnegative reals $[0,\infty)$, because we can view this space as  $\omega$-many copies of $[0,1)$:
$[0,1)[1,2)[2,3)...$
Your question about the ant is related to path connectedness (which is equivalent to connectedness here).  There is no continuous path from a point in $(n-1,n)$ to a point in $(n,n+1)$ as the subspace $(n-1,n)(n,n+1)$  is not connected, that is, it is the union of two nonempty open subsets. In the reals this would not be a problem as $(n-1,n+1)$ is connected.
A: The relevant concept from topology here is connectedness. You may want to read the Wikipedia article for more information. There are lots of equivalent ways of defining it, but a topological space is considered if the space can't be divided into two open sets. The terms "open" and "closed" are misleading because sets can be both open and closed, or neither, but when your topological space is a subset of the real line using the Euclidean topology, then it makes sense because the open sets are unions of open intervals, and the closed sets are (finite) unions of closed intervals. Here are a few examples:
$(0, 1)$ is open.
$[0, 1]$ is closed.
$(0, 1) \cup (1, 2)$ is open.
$(0, 1) \cup [1, 2]$ is neither open nor closed. This is example is actually a single interval, but it doesn't matter.
So, to answer your question about the set $\mathbb{R} - \mathbb{Z}$ using the euclidean topology, it's easy to show that it's not connected. The set of all points less than $0$ and the set of all points greater than $0$ are two disjoint open sets such that their union is the whole space.
Another counterexample is to have one open set be the interval $(0, 1)$ and the other set be everything else. This shows that the points $0.5$ and $1.5$ are in connected components, so there's no path from one to the other with the usual topology.
A: From The Continuous and the Infinitesimal by John L Bell, p280: 'Taken together, these principles have been shown to have remarkable consequences for the indecomposability of subsets of the continuum. Not only is the intuitionistic continuum indecomposable, but, assuming the Continuity Principle and Kripke's Scheme it remains indecomposable even if one pricks it with a pin. [More exactly, for any real number a, the complement R - {a} of {a} is indecomposable.] "The [intuitionistic] continuum has, as it were, a syrupy nature, one cannot simply take away one point." [In Van Dalen (1997) the classical continuum is described as the "frozen intuitionistic continuum".] If in addition Bar Induction is assumed, then, still more surprisingly, indecomposability is maintained even when all the rational points are removed from the continuum.'  
