# Diagrams of topological terms

1. Jan 11, 2010

### beetle2

Hi Guy's
I am just starting out in topology and I was wondering if someone might know of a good link that may have venn diagrams of some important topological terms ie closure of A, int A, limit points etc.

regards
Brendan

2. Jan 11, 2010

### HallsofIvy

I can't imagine Venn diagrams as being a good way to keep those straight. They would get much too complicated. Just learn the definitions.

3. Jan 11, 2010

### beetle2

For A = (0,1). a limit point P of A is that P is a point such that every open set around it contains at least one point of A different from P.

So for this example 0 and 1 would be limits points of A although they are not in A, and there would be infinitley many limits points of A.

So if I take 0 which is not an element of A and an open set around it (0-e,0+e) the point 0+e is in A and is not equal to 0. hence a limit point
The same would be for 1.

Take 1 which is not an element of A and an open set around it (e-1,1+e) the point e-1 is in A and is not equal to 1.

4. Jan 11, 2010

### beetle2

If I have $$X = (- \infty,0] \cap [1,\infty +)$$ and $$A \subset X = (0,1)$$

would A = (0,1) be the interior of A and the closure of A = $$X\bar{A}$$?

5. Jan 12, 2010

### HallsofIvy

This makes no sense. They way you have defined X, it is the empty set- there is NO real number that is in both $(-\infty, 0]$ and $[1, +\infty)$, those two sets are disjoint. At first I thought you meant "$\cup$" rather than "$\cap$ but the rest would still make no sense. You have defined X as "all real numbers except (0, 1) so "$A\subset X= (0, 1)$" is nonsense. With either $cup$ or $cap$, X is NOT equal to (0, 1). If you mean A= (0, 1), then A is not a subset of X. If you meant "$\cap$", X is empty and has only itself as subset. If you meant $\cup$, A is, in fact, the complement of X.

If A is (0, 1) as a subset of the real numbers, with the usual topology, then its interior is (0, 1) (A is open) and its closure is [0, 1]. If you meant X as the underlying set with the topology inherited from the real numbers, whether you meant "$\cap$" or "$\cup$", A is not a subset of X.

6. Jan 12, 2010

### beetle2

Thanks for you reply I see where I stuffed up. I did mean...
$$X= R$$
$$A \subset X \mid x\in (0,1)$$

If A is (0, 1) as a subset of the real numbers, with the usual topology, then its interior is (0, 1) (A is open) and its closure is [0, 1].

I was confused how the closure of A was [0,1] then I re-read the definition in my text.

It says.

A point x is in the closure of $$A$$ if for each neighbourhood $$N$$ of x $$N \cap A = \emptyset$$

So if I take x=0 which is not an element of A and an neighbourhood N around it say (0-e,0+e) the point $$0+ \eps \in A \cap N$$ the same would be for N = (1-e,1+e) $$1- \eps \in A \cap N$$

regards