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Are $\{0\},\{1\}$ clopen in $\{0,1\}$ with the Euclidean metric?

I think they are, but I would like a confirmation. Thank you.

YoTengoUnLCD
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2 Answers2

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If you are using the Euclidean metric as you indicated in the comments then yes.

$N_{.25}(0) = \{x \in \{0,1\}| d(x,0) < .25\} = \{0\}$ so {0} is open. Similar for {1}. As closed means compliment is open, $\{0\}^c = \{1\}$ which is open so {0} is closed. Similar for {1}.

But to be more general to declare a set open and closed one has to declare a topology on the space. This is, more or less, arbitrarily defining certain sets to be open by fiat. {0,1} as a subset of real numbers (or any other space) with the euclidean (or any other) metric is simply one arbitrary topology. We could just as easily declare a topology where neither {0} nor {1} are open or closed. (Although in all topologies $\emptyset$ and the universal set are open. For a set with two elements there are four possible topologies on whether we declare which of the proper subsets to be open or not.)

(So the four possible topologies are:A {0} and {1} are not open and not closed. This is called the trivial topology. B: {0} is open but {1} isn't. So {1} is closed but {0} isn't. C: {1} is open but {0} isn't. So {0} is closed by {1} isn't. D: both {0} and {1} are open. So both are closed. This is equivalent to any metric one can place on {0,1}.)

(Postscript: If {a,b} is any metric space, not just the euclidean, nor anesc. a subset of the reals, then $c = d(a,b) > 0$ so for any $0 < \epsilon < c$ we have $B_{\epsilon}(x) = \{y \in\{a,b\}| d(x,y) < \epsilon\} = \{x\}$ so {x} is open. So in any 2 element metric space all sets are clopen. Actually in all finite metric spaces all sets are clopen.)

fleablood
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  • I dont have completely clear that a distance of $0.25$ is legitimated for a set as ${0,1}$ where only exist two kind of distances: zero or one. But the result is the same if we take 1 instead of 0.25. – Masacroso Apr 18 '16 at 19:35
  • Any non-negative real number can be a potential distance. The fact that there aren't any points where d (x,y)= .25 doesn't matter. Actually that there aren't any* is the point. The set of all points x where d (0,x) are less* then .25 are precisely the same set as all points x where d (0,x) =0 because 0 is the only possible distance less than .25. If you want, you can do a neighbor hood of less than 1, as d (xy)< 1 iff d (x,y)=0 iff x=y. – fleablood Apr 18 '16 at 20:57
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Yes. Both sets are clopen. To see this note that $\{0\}=B(0,\frac{1}{2})$ and $\{1\}=B(1,\frac{1}{2})$, both of which are open sets. So, $\{0\}$ and $\{1\}$ are both open. They are also both closed since they are complements of each other.

ervx
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