Let $\mathcal S$ be the collection of all straight lines in the plane $\mathbb R^2$. If $\mathcal S$ is a subbasis for a topology $\mathcal T$ on the set $\mathbb R^2$, what is the topology?

I know that the topology must be the discrete topology, but I can't seem to think how I can go about showing it. I tried searching online and came across this (see picture below), but something does not seem correct about it:

enter image description here

The above screenshot was taken from here.

The reason I feel that this is not correct, is because they use the fact that the point $(a,b)$ is open. However, $\mathbb R^2$ is Hausdorff, and so singleton sets (i.e points) are closed?

Is there any other way that I can go about proving this?

  • 2
    $\begingroup$ $\mathbb R^2$ is Hausdorff with the standard topology. We are imposing a new topology generated by $\mathcal S$, thus it need not be Hausdorff. $\endgroup$ – Trevor Norton Apr 27 '16 at 20:53
  • $\begingroup$ @TrevorNorton . That makes sense :). Thanks! But how can we conclude that the point $(a,b)$ is in fact open in the topology generated by $\mathcal S$? $\endgroup$ – user290425 Apr 27 '16 at 20:55
  • $\begingroup$ Every set in the subbasis is open, and any finite intersection of sets from the subbasis is also open. Since we can pick two lines whose only intersection is a single point (say $x=a$ and $y=b$), we can say that the single point is also open. $\endgroup$ – Trevor Norton Apr 27 '16 at 20:58
  • $\begingroup$ Actually, this new topology is also Hausdorff ... $\endgroup$ – Hagen von Eitzen Apr 27 '16 at 21:01
  • $\begingroup$ @HagenvonEitzen . Does the argument given by Trevor above still hold though in showing that the point $(a,b)$ is open? $\endgroup$ – user290425 Apr 27 '16 at 21:11

Let $\mathscr T$ be a topology on a non-empty set $X$ and let $\mathscr S\subseteq \mathscr T$. Then, by definition, $\mathscr S$ is a subbasis for the topology $\mathscr T$ if for any $U\subseteq X$, one has that $U\in\mathscr T$ if and only if $U$ can be expressed as a union of sets which are finite intersections from sets in $\mathscr S$. Formally: $$U=\bigcup_{\gamma\in\Gamma}\bigcap_{i\in F_{\gamma}} S_i^{\gamma},$$ where

  • $\Gamma$ is an (arbitrary) index set,
  • for each $\gamma\in\Gamma$, $F_{\gamma}$ is a finite index set, and
  • for each $\gamma\in\Gamma$ and $i\in F_{\gamma}$, it holds that $S_i^{\gamma}\in\mathscr S$.

In the concrete example, note that any set consisting of a single point in $\mathbb R^2$ can be expressed as the intersection of two lines, which lines are in $\mathscr S$ by assumption (in this case, the index set $\Gamma$ has a single element $\gamma$, and the index set $F_{\gamma}$ has two elements, corresponding to the two lines). This implies that every one-point set in $\mathbb R^2$ is open according to this new topology $\mathscr T$, so that $\mathscr T$, in fact, must be the discrete topology (in which every subset is open).

In the light of your question:

However, $\mathbb R^2$ is Hausdorff, and so singleton sets (i.e points) are closed?

Two comments: Firstly, the new topology $\mathscr T$ is different from the standard Euclidean topology. This is a common source of confusion, since we are so used to working with the usual Euclidean topology on $\mathbb R^2$ that we have a strange feeling upon seeing that nothing prevents us from defining any other topology on $\mathbb R^2$. Therefore, that the Euclidean topology is Hausdorff does not say anything about whether the discrete topology $\mathscr T$ is Hausdorff or not.

That said, secondly, the discrete topology on $\mathbb R^2$ does happen to be Hausdorff, so that it is also $T_1$ (i.e., the singletons are closed). This is compatible with the fact that singleton sets are open in the discrete topology, since any subset of the space is both open and closed under this topology.


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