General Topology – Proof of the Usual Topology on R

analysisgeneral-topology

I'm trying to prove that the usual topology on R is in fact a topology for R. I've read through a few books on topology and most of them seem to skip over the proof and list it as "trivial". My math skills are obviously lacking because I'm stumped! I'm trying to prove the following three things:

  1. That R and the empty set are elements of the usual topology
  2. That the union of any two elements is an element of the usual topology
  3. That the intersection of any two elements is an element of the usual topology

Of these, 3 to me seems the easiest to prove because if two elements don't overlap, the answer is the empty set, otherwise it is given by the interval that the two elements overlap.

I'm having a few issues with 1 and 2 though. Specifically:

Why is the empty set in the usual topology? I can't find a value x in R or e > 0 such that the empty set = x +- e?

Likewise, why is R in the usual topology? For example, take x = 0 and e = (abstractly) "the biggest number in R". Then we have the interval from -"the biggest number" to +"the biggest number", but by definition that cannot include "the biggest number" so isn't R?

By this logic, it seems that maybe a better definition for the usual topology is not only all intervals in R, but also R and the empty set as well?

Lastly, I'm struggling to prove that the union of two elements is also in the usual topology. Consider the intervals (1, 2) and (3, 4). The union of these two isn't an interval, hence shouldn't be a part of the usual topology?

Now obviously several centuries worth of mathematicians are going to be vastly more correct than me (and that's something I can actually prove!) so my big question is… where am I going wrong? Is it in my understanding of the usual topology, of topologies in general or in my logic?

Thanks!

EDIT: Sorry, I should have mentioned the definition of "usual topology". The usual topology T is defined such that that O is an element of T if and only if for all x in O there exists an e > 0 such that the interval (x – e, x + e) is a subset of O.

Best Answer

First a reminder of the definition of a topology:

Definition 1: A collection $\tau$ of subsets of some set $X$ is called a topology on $X$, iff

  • $\emptyset, X\in \tau$

  • An arbitrary union of sets in $\tau$ is again in $\tau$.

  • A finite intersection of sets in $\tau$ is again in $\tau$.

Now your other mistake is that you are not making yourself aware of how the usual topology on $\mathbb{R}$ is actually defined:

Definition 2: A set $U\subset\mathbb{R}$ is called open iff for every $x\in U$ there exists an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subset U$.

So for instance, $\emptyset$ is really open by that definition, since it does not have any elements $x$ for which there would have to exist an $\epsilon>0$ such that (..).

Also, $\mathbb{R}$ is tautologically open by that definition, since the sets $(x-\epsilon,x+\epsilon)$ are always contained in $\mathbb{R}$, no matter what $x,\epsilon\in\mathbb{R}$ are exactly.

Now let us take an (arbitrarily large, possibly uncountable) index set $I$ and for every $i\in I$ an open set $U_i\subset\mathbb{R}$. Consider the set

$$U=\bigcup_{i\in I} U_i$$

We have to show that $U$ is open. Therefore take some $x\in U$, but then $x\in U_i$ for some $i\in I$. But $U_i$ is open, hence (.. can you finish the thought?)

To check that finite intersections of open sets are again open, it suffices to check this for the intersection of two open sets. Let $U,V\subset\mathbb{R}$ be open. We claim that then $$W=U\cap V$$ is open as well.

So let $x\in W$. Then $x\in U$ and $x\in V$. But $U$ and $V$ are open, hence there exists (.. can you finish it from here?).

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