Let $(X, d)$ be a metric space where $X$ is the set of real numbers and $d$ is the discrete metric. Is the subset of this metric space $A=\{x \in \mathbb{R}:0 < x < 1 \}$ closed?

Here is what I have so far.

A set is closed if it contains all of its limit points. So this problem is really asking if A contain all of its limit points.

This leads me to ask, what are the limit points of A?

I think I have a good grasp of the concept of a limit in the context of a sequence in a metric space: if you pick a distance, $\epsilon$, I can give you a natural number, N, where the distance between any two terms of the sequence after $S_N$ is less than your $\epsilon$.

The more I think about it, I am not really sure how to think about a limit in the context of a set. My gut sense is that the only two limits are 0 and 1, since the set gets arbitrarily close to 0 and arbitrarily close to 1. Though my gut feeling is also telling me that this is wrong.

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    $\begingroup$ If a set has the discrete topology then every subset is open, hence every subset is closed. $\endgroup$ Sep 15 '16 at 22:19
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    $\begingroup$ The set does not get arbitrarily close to $0$ or $1$. All points are distance $1$ from one another in the discrete metric. Think about that when doing your $\epsilon$ argument with, for instance, $\epsilon=0.5$. $\endgroup$
    – Arthur
    Sep 15 '16 at 22:31
  • $\begingroup$ Limit points in the discreet metric are impossible. They do not exist. Every open neighborhood of radius 1 or less will contain only the one center point and no other. Thus no point can be a limit point for any set. $\endgroup$
    – fleablood
    Sep 15 '16 at 23:00
  • $\begingroup$ O and 1 are not limit points. B(0, 1) = {0} and {0} has no points in A.. Ditto B(1,1) = {1} has no points in A. $\endgroup$
    – fleablood
    Sep 15 '16 at 23:02

In the discrete topology, the only type of convergent sequence is a constant sequence, for if the sequence is non-constant set $\epsilon=1$ and observe that the sequence will never get within $\epsilon=1$ of its proposed limit point.

Now, this depends on your definition of limit point - but if you allow constant sequences, then every point is a limit point of its own constant sequence.

To answer the specific question, however, observe that every singleton set is open, because setting $r<1$, $N_r(x)=\{x\}$ for any $x\in X$. Then because every $A\subset X$ is a union of its singleton subsets, we can see that every set $A$ is open in $X$. Thus, every set is closed, for its complement is always open.

So, the set you're wondering about is closed.

  • $\begingroup$ Thanks, I follow the first part of what you are saying (the only convergent sequence in a discrete metric space is a constant sequence). I don't understand why every singleton set is open though. $\endgroup$ Sep 16 '16 at 0:11
  • $\begingroup$ Is there any part that needs clarification? $\endgroup$ Sep 16 '16 at 0:13
  • $\begingroup$ Yes, why exactly is a singleton set open? $\endgroup$ Sep 16 '16 at 0:17
  • $\begingroup$ In the standard definition of a metric topology, a $U$ is open if $\forall x\in U$ there exists $r>0$ such that $d(x,p)<r$ implies $p\in U$. In this case, it is impossible for $p\ne x$ to satisfy $d(x,p)<1$, by definition of the discrete metric. Let a singleton set $\{x\}$ be given, if we choose $0<r<1$, then $d(x,p)<1$ implies $x=p$, and so we have checked that for each $y\in \{x\}$, there exists an $r>0$ such that $d(y,p)<r$ implies $p\in \{x\}$. In particular, the only $y$ here is $y=x$. $\endgroup$ Sep 16 '16 at 0:19
  • $\begingroup$ Small correction: the only convergent sequences are the eventually constant ones. For instance, $(1,2,3,3,3,\ldots)$ is not constant, but converges to $3$. $\endgroup$
    – Arthur
    Sep 16 '16 at 5:34

I think I would prefer a proof based on the fact that $X'$ is open. But since you want to look at limit point, lets look at limit points.

X is a closed set. Every limit point of X is in X.

Proof by contradiction.

Suppose, y is a limit point of X and y is not in X.

Then there exits an x in X such that for any $\epsilon > 0, d(x,y) < \epsilon$

Let $\epsilon = \frac 12$

$d(x,y) = 1 > \epsilon$ , Contradiction!

There are no limit points of X that are not in X.


"Let (X,d) be a metric space where X is the set of real numbers and d is the discrete metric. Is the subset of this metric space << mumble-muble--- I got distracted by a cute cat >> closed?"

Answer, wait what subet?.... Um, YES. Yes it is closed because all subsets are closed.

Let $S$ be .... um, whatever it was you said, I wasn't paying attention.

Is $y$ be a limit point of $S$. Then every open neighborhood of $y$ contains a point in $S$ that is not equal to $y$. So the the open neighborhood $B(y, 1/2) = \{x\in X| d(x,y) < 1/2\}$ contains a point of $S$ that is not equal to $y$.

But If $d(x,y) < 1/2 \iff d(x,y) = 0 \iff x = y$. So $B(y, 1/2) = \{x\in X| d(x,y) < 1/2\}= \{y\}$. So $B(y, 1/2)$ does not contain any point of $S$ not equal to $y$. It doesn't contain any points at all not equal to $y$. So $y$ is not a limit point.

No sets in $X$ with the discrete metric spaces have limit points at all! In the discrete metric space there is no such thing as limit points!!!!

So ... well, all sets are vacuously closed.

So yes, S, is closed. I don't even know what $S$ is, but it is closed.


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