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I learned recently in class that Q is not open in R because any subset of Q contains irrationals between the two points. So you cannot fit an open ball inside Q.

What about Q intersected with (a,b). Is this set relatively open in Q?

How much of a role does the actual metric space play in determining whether a set is open or not? If we take the metric space to be Q, and consider the set Q. Then Q is open in this space compared to it being closed in the metric space R.

I also would like to consider the set {0,1}. This set was recently in a proof I looked over and they mentioned that {0} was open, and {1} was also open. I can't see how {0} would be open if we assumed that {0,1} was in the metric space R. How could we fit an epsilon ball centered at 0 inside the set {0,1} since it only has two elements?

I can see if {0,1} was considered to be its own metric space, and with the normal metric, just take any epsilon ball centered at 0 with epsilon less than 1, then it would be completely contained in {0,1}.

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The first thing you should note is that whenever someone claims that a set $U$ is open, you should always ask "open in what?".

The answer to your first question is what is called the subspace topology. That is, given a topological space $X$ and a subset $Y\subset X$, we can define a topology on $Y$ by declaring $U'\subset Y$ as open precisely when it has the form $U\cap Y$ for $U$ an open set of $X$ in the original topology. So your set $(a,b)\cap \mathbb Q$ is open in $\mathbb Q$ with the subspace topology induced from $\mathbb Q\subset \mathbb R$. So the actual topology of the original space (and so the metric which induces it) plays a direct role in the subspace topology.

Again if I take the set $Y=\{0,1\}\subset \mathbb R $ with the subspace topology, clearly $\{0\} = (-0.5,0.5)\cap Y$ so is open in the subspace toplogy. Similarly with $\{ 1\}=(0.5,1.5)\cap Y$ is also open.

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  • $\begingroup$ Thank you. Things are starting to become clearer. When you mention that (a,b) intersect Q is open in Q. Does it inherit the same properties as open sets. Such as for every element in (a,b) intersect Q, there is a open ball centered around that element inside (a,b) intersect Q. I see how {1} is considered open, but there doesn't seem to be a way to have an epsilon ball centered at 1 as a subset of {1} $\endgroup$ – dfk3 Oct 21 '17 at 19:05
  • $\begingroup$ Indeed, just find the desired open ball in $\mathbb R$ and intersect it with $\mathbb Q$. The metric structure is inherited mutatis mutandis. Yes you do, consider the open ball of radius $0.5$ centered around $1$ in $\mathbb R$. When you intersect this with $Y=\{0,1\}$ you will notice that the open ball of radius $0.5$ around $1$ is exactly the single point $1$. $\endgroup$ – Ravi Oct 21 '17 at 19:09
  • $\begingroup$ Oh, ok I think I got it! So {1} and {0} are open in the subspace topology. But they aren’t open in R. So just because a set is open in a subspace topology doesn’t guarantee that it’s open in actual topology. And vice versa. $\endgroup$ – dfk3 Oct 21 '17 at 19:47

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