Understanding Open Sets in Real Numbers

In the context of real analysis and topology, identifying an open set is a fundamental concept. A set \(S\) in \(\mathbb{R}\) (the set of real numbers) is considered an open set if, for every point \(x\) in \(S\), there exists a small open interval \((x - \epsilon, x + \epsilon)\) (where \(\epsilon > 0\)) that is entirely contained within \(S\). This means that you can move a tiny bit in any direction from any point within the set and still stay inside the set.

Analyzing the Given Options for Open Set Properties

Let's examine each given set to determine if it fits the definition of an open set.

• Option 1: \( S = \left\{ {1,\dfrac{1}{2},\frac{1}{3},...} \right\}\)

  This set consists of isolated points. Consider the point \(1 \in S\). If we take any open interval around 1, say \((1-\epsilon, 1+\epsilon)\) with \(\epsilon > 0\), this interval will contain infinitely many points that are not in \(S\) (for example, \(1 + \epsilon/2\) is in the interval but not in \(S\)). Since we cannot find an open interval around 1 that is fully contained within \(S\), this set is not an open set.

• Option 2: \( S = \mathbb{N}\) (The set of natural numbers)

  The set of natural numbers is \( \{1, 2, 3, ...\} \). Similar to the previous option, these are isolated points on the real number line. For any natural number \(n \in \mathbb{N}\), any open interval \((n - \epsilon, n + \epsilon)\) for \(\epsilon > 0\) will contain non-natural numbers. For instance, \(n + \epsilon/2\) is in the interval but not in \(\mathbb{N}\). Thus, \(\mathbb{N}\) is not an open set.

• Option 3: \( S = \mathbb{Z}\) (The set of integers)

  The set of integers is \( \{..., -2, -1, 0, 1, 2, ...\} \). Like natural numbers, integers are isolated points. For any integer \(m \in \mathbb{Z}\), any open interval \((m - \epsilon, m + \epsilon)\) for \(\epsilon > 0\) will contain non-integer points. For example, \(m - \epsilon/2\) is in the interval but not in \(\mathbb{Z}\). Therefore, \(\mathbb{Z}\) is not an open set.

• Option 4: \( S = \{x \in \mathbb{R} : 1 < x < 3\}\)

  This set represents the open interval \((1, 3)\) on the real numbers line. Let's take any point \(x\) in this set, so \(1 < x < 3\). We need to find an \(\epsilon > 0\) such that the interval \((x-\epsilon, x+\epsilon)\) is completely contained within \((1, 3)\). We can choose \(\epsilon\) to be smaller than the distance from \(x\) to 1 and also smaller than the distance from \(x\) to 3. That is, \(\epsilon < x - 1\) and \(\epsilon < 3 - x\). A simple choice is \(\epsilon = \min(x-1, 3-x)/2\). With this choice of \(\epsilon\), the interval \((x-\epsilon, x+\epsilon)\) will be entirely within \((1, 3)\). Since we can do this for every point \(x\) in \(S\), this set \(S = (1, 3)\) is an open set.

Conclusion

Based on the analysis of each option and the definition of an open set, only the set representing the open interval \((1, 3)\) satisfies the condition that every point within it has a neighborhood entirely contained within the set.