# Is the set $\left\{x \in Q\colon x^2 \le 2\right\}$ open or closed?

Is the following set open or closed? I am almost certain it is open as the limits are not included in the rational set.

$$\left\{x \in \mathbb{Q}\colon x^2 \le 2\right\}$$

What I really don’t understand is the proper closure of the set. I think it would be:

$$\left\{x \in \mathbb{Q}\colon x^2 \le 2\right\} \cup \left\{\pm\sqrt{2}\right\}$$

But then again, the following seems reasonable (although not an “efficient” closure):

$$\left\{x \in \mathbb{R}\colon x^2 \le 2\right\} = \left[-\sqrt{2},\sqrt{2}\right]$$

• A set if open as a subset of a set with a topology. Which one is the super set with topology that is supposed to contains those sets? $R$? $Q$? – OR. Jun 18 '17 at 2:31
• Take the first set. As a subset of $Q$, considering $Q$ with the topology induced by $R$, we get that the set is open, since $A:=\{x\in Q:\ x^2\leq2\}=Q\cap(-\sqrt{2},\sqrt{2})$. The right-hand side is an open set of $Q$ by definition. On the other hand, as a subset of $R$ the set $A$ is neither open nor closed. It is not closed, as you were saying for having limit points outside itself. It it not open because no neighborhood of $0$ is completely contained in $A$. – OR. Jun 18 '17 at 2:41
• "I am almost certain it is open as the limits are not included in the rational set." That would only mean it is not closed. It wouldn't mean it is open. It could very well be neither. – fleablood Jun 18 '17 at 21:03
• "Is the following set open or closed?" Why do you assume it has to be one or the other? Could it be both? Could it be neither? – fleablood Jun 18 '17 at 21:04

Let $A$ be your set of rationals. You ask if $A$ is open or closed.

Open or closed within what space? The reals $\mathbb{R}$ or the rationals $\mathbb{Q}$?
Notice that $$A = [-r,r] \cap \mathbb{Q} = (-r,r) \cap \mathbb{Q}$$ where r = sqrt 2. Thus within $\mathbb{Q}$, $A$ is clopen.
However, within $\mathbb{R}$ it is neither.

The $\mathbb{R}$-closure of $A$ is $[-r,r]$. The $\mathbb{R}$-interior of $A$ is empty.

It depends on what space your topology is.

If your space is $\mathbb R$ it is neither open nor closed.

If your space $\mathbb Q$ it is both open and closed.

This set is simply $[-\sqrt{2}, \sqrt{2}]\cap \mathbb Q$.

The limit points are all the points between $-\sqrt{2}$ and $\sqrt{2}$. In the space $\mathbb Q$ that is indeed the set. So in the space $\mathbb Q$ this set is closed.

But in the space $\mathbb R$ this includes all the irrational points as well. The rational limit points are included but the irrational limit points are not. So in the space $\mathbb R$ this set is not closed.

In the space $\mathbb Q$ every point is an interior point. Around every point you can find a small enough neighborhood that is entirely inside the set. So every point is an interior point. So the set is open in $\mathbb Q$.

But in the space $\mathbb R$ every neighborhood no matter how small will include irrational points that are not part of the set. So no point of the set is actually an interior point. So the set is not open in $\mathbb R$.

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In $\mathbb Q$ the set is closed. So the closure is the set itself.

In $\mathbb R$ the set is not closed. To get the proper closure you must include all the limt points. $\pm \sqrt{2}$ are limit points, yes, but so is every point between $-\sqrt{2}$ and $\sqrt{2}$. A limit point isn't just "the points on the edge". There are the points so that every neighborhood no matter how small will intersect the set. These are not only the points "on the edge" but the points "inside" as well.

So the closure will include all the points between $-\sqrt{2}$ and $\sqrt{2}$ and so the closure in $\mathbb Q$ is the set $[-\sqrt{2}, \sqrt{-1}]$.

===

Note:

Open $\ne$ not closed. Sets can be both and sets can be neither.

limit points $\ne$ boundary points. boundary points $\subset$ limit points.

It is open in $\Bbb Q$ as you correctly observe. It is closed in $\Bbb Q$ because it is the inverse image of a closed interval under the (continuous) squaring function.