Re: Connectedness question

none wrote: 
Connectedness is (in this example) a _topological_ notion.

The phrase "R is connected" actually means "if R is equipped with the
usual topology, then the resulting topologal space is connected".

[...]

Just review the definitions of "connected" and "complete" for topological (or metric spaces(. 


Ok, but isn't the topological notion of connectedness _chosen_ as it is
 because we want R to be connected and Q no to be (with the usual
topology)?


Topologically, a space is connected if it is not the disjoint union of
nonempty open subsets.  A priori, this doesn't have anything to do with
either R or Q; the reason that R ends up connected is a coincidence from
this perspective, in that it relies on:

1) R being a densely ordered field;
2) the topology being the order topology;
3) the topology also being metrizable, and the metric a complete one.

Incidentally, the only ordered field in which the order topology is
metrizable and the metric is complete is R.  Therefore, if one takes the
perspective you are taking, that R is the canonical example of a connected
space, then one is led to the definition that ANY topological space should
be connected if and only if two points lie in a continuous image of the
real line.  That's what we call path connected now.  The standard,
abstract definition is more useful for doing proofs in many cases, is
equivalent in any space people consider "reasonable", and does not rely on
the existence of a complete ordered field.

I mean, the intuition for the concept of connectedness when it was defined in topology came from experience with R and Q. But this experience was arbitrary since in another conceivable world R would not
be considered "connected" (before the notion was rigorized). 

The surreals are not actually a set so we'll pretend you're talking about
the "hyperreals", where one adds infinitesimals in a more controlled
fashion.  It's the same thing, morally, in this conversation.  The problem
is that Q is disconnected because its topology is the _subspace_ topology
of R, and we have removed points which, since the topology is the order
topology, disconnect Q by definition.  The topology on R is _not_ the
subspace topology of the hyperreals.  Take this example: pick an
infinitesimal which is greater than 0 but less than every positive real
number.  It defines two open sets, of everything on one side or the other
of it.  Look at the one on the left: it intersects R in the _nonpositive_
real numbers, which is _not_ an open set.  Therefore you can't use the
same "full of holes" argument to disconnect R as you could for Q.

I guess I lack some understanding of basic topology :/ 

No, this is a good question.

--
Ryan Reich
ryanr@xxxxxxxxxxxx
.

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