Suppose that we have a set $S$ containing 0 and 1. Can we define our topology to be the four open sets $\varnothing$, $\{0\}$, $\{1\}$ and $\{0,1\}$? I know that the Sierpinski set contains the three elements $\varnothing$, $\{0\}$, and $\{0,1\}$. I wonder if the fourth element is added, is it still a topology.

  • 1
    $\begingroup$ There are really just three axioms to check, and only four sets to check them with. Even if you do everything by hand, completely robotic, without optimization, it shouldn't take you longer than two hours to check that. And if you do it using your brain, then it should take only a couple of minutes. $\endgroup$
    – Asaf Karagila
    Commented Oct 24, 2014 at 18:10
  • $\begingroup$ I don't get it, unless $S = \{0,1\}$ it won't be. Just saying $S$ containing 0 and 1 obviously doesn't cut it. $\endgroup$
    – dioid
    Commented Oct 24, 2014 at 18:35
  • $\begingroup$ @dioid: That appears to be OP's intent. $S=\{0,1\}$. $\endgroup$
    – MPW
    Commented Oct 24, 2014 at 19:26

1 Answer 1


$\mathscr P(S)$ is a topology on $S$ for any set $S$, often called the discrete topology.

In this topology, every subset of $S$ is open, so every map defined on $S$ is continuous. Pretty simple, but extremely useful.

  • $\begingroup$ it should be a comment ! $\endgroup$
    – idm
    Commented Oct 24, 2014 at 18:20
  • $\begingroup$ @idm: What do you mean? How does this not answer the question? $\endgroup$
    – MPW
    Commented Oct 24, 2014 at 18:21
  • $\begingroup$ I did see it before it got corrected. I'll remove the previous comment. $\endgroup$
    – paw88789
    Commented Oct 24, 2014 at 18:22
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    $\begingroup$ It may be boring, but it's danged useful for disproving a bunch of things, like "do continuous functions preserve quality X". Any quality that the discrete topology has automatically fails that test, since you can always define a continues function from a space under the discrete topology to the same space under the actual topology as the identity map $\endgroup$
    – Alan
    Commented Oct 24, 2014 at 18:37
  • $\begingroup$ $\mathscr P(S)$ is a topoly on $S$ but it only contains the four open sets stated in OP of $S$ unless $S$ contains any other elements. And unless that is the case the four elements won't be a topology. $\endgroup$
    – dioid
    Commented Oct 24, 2014 at 19:18

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