Science Forums 2

Ben · 06-15-2009

Quote:

Originally Posted by Qfwfq

Wait a minute, I protest! A neighborhood is not necessarily open!

Well, I grant you 2 things; first my definition was rather imprecise, and second there seems to be no agreement on this in the literature.

Let's go with this compromise, which I copy from Bishop & Goldberg Tensor analysis on Manifolds

Quote:

Originally Posted by Bishop & Goldberg

A neighborhood of $x \in X$ is any $A \subsetneq X$ such that $x \in A^o$ . In particular, any open set containing $x$ is a neighborhood of $x$

(my bolding).

I will happily concede my original definition was wrong iff it can be shown that, for any pair, say, of closed sets $A \ne B$ that $A^o = B^o$ . I doubt it, but I am not completely sure. Someone help me out here!

Quote:

In topology, so many roads lead to Rome.

How true.

While I am here, I promised this

Quote:

Originally Posted by me

A topological space that cannot be written as the union of 2 non-empty disjoint sets is said to be connected.

Alternatively, in a connected space, the only sets that are both open and closed are the base set and the empty set, say $S$ and $\O$ . It is a fun exercise to bring these two definitions into register.

So, here it is.

Suppose that $X$ is connected by our second definition. Let $A \subsetneq X, \, B \subsetneq X$ be open. Let $X = A \cup B$ , and let $A \cap B = \O$ (recall this is the the definition of disjointness).

Then of necessity, $B = A^c \in X,\, A = B^c \in X$ , therefore $A$ and $B$ are also closed, which by our first definition means that $X$ is not connected, a contradiction, and therefore our two definitions coincide.

PS Yikes!! I am losing my marbles, this not what I promised at all. Recall I defined the boundary of an arbitrary set $A$ by $\partial A = A^- \setminus A^o$ .

If $A$ is both open and closed, then $A = A^- = A^o \Rightarrow \partial A = A^- \setminus A^o \equiv A \setminus A = \O$
And what if a set is neither open NOR closed?

Exercise for readers.

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