2

In the definition of topology:

Any union of elements of $τ$ is an element of $τ$ and any intersection of finitely many elements of $τ$ is an element of $τ$

Why for union, INFINITE union of open sets is open while in case of intersection, FINITE intersection of open sets is open?

Thank you.

mvw
  • 34,562
  • 3
    Open sets are meant to give is wiggle room. If we take unions of things that have wiggle room, we will have wiggle room. But if we take the intersection of things that have increasingle small (but existent) wiggle room, we may end up with none. For example, in the real line $\bigcap_{n\geqslant 1} (x-1/n,x+1/n)={x}$ – Pedro Feb 02 '15 at 14:48
  • @Pedro Tamaroff: What is 'wiggle room'? –  Feb 02 '15 at 14:53
  • All of the open intervals $(x-1/n, x+1/n)$ contain the point $x$. – mjqxxxx Feb 02 '15 at 14:54
  • Relevant: http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets – Hayden Feb 02 '15 at 14:56
  • 1
    What wiggle room means depends on the space. In a metric space it means that for any $ x$ in the set, we will remain in the set provided we do not go more than some distance $ \delta (x) $ from $ x$. The interpretation is similar in general spaces. – Ian Feb 02 '15 at 15:00

2 Answers2

1

Intuitively, a set $U$ is "open," roughly, if, for any $x\in U$, some neighborhood of $x$ is in $U$.

This is why the arbitrary union of open sets is open. If $x\in U=\bigcup U_i$ then $x\in U_i$ for some $i$, and some neighborhood of $x$ is in $U_i$, and hence that neighborhood is in $U$.

This is not true for arbitrary intersections.

Thomas Andrews
  • 177,126
1

The definition of a topology was constructed so that it could satisfy the Euclidean topology on $\mathbb R$ but would be flexible enough to include many other spaces.

Our notion of openness on the real line is that given an open set $E$ (in the metric sense) that every point inside it has a ball containing the point but also contained in $E$. We know that taking an arbitrary union of these sets preserves this property, but isn't so for infinite intersections. As mentioned in the comments above, if we take an infinite intersection of sets of the form $\left ( -\frac{1}{n}, \frac{1}{n} \right )$ then we are left with the set $\{0\}$ which we don't regard as open since every open ball around 0 necessarily contains points outside the singleton set containing 0. The definition of a topology extends this same idea to other spaces that may not have a metric associated to it.

Mnifldz
  • 12,880