Exercise on Limit Points in Topology 4

In this exercise, I need to determine the set of limit points of \( A = (-1, 1) \cup \{2\} \) in the standard topology on \(X=\mathbb{R}\).

In the standard topology on \(\mathbb{R}\), open sets are unions of open intervals.

To find the accumulation points of the set \( A = (-1, 1) \cup \{2\} \), I need to identify which points \( x \) in \( \mathbb{R} \) have a neighborhood that contains at least one point of \( A \) other than \( x \).

• Interval \((-1, 1)\)
  Consider a point \( x \in (-1, 1) \). Every open neighborhood of \( x \) in \(\mathbb{R}\) is of the form \((x - \epsilon, x + \epsilon)\) with \(\epsilon > 0\) and always contains other points of \((-1, 1)\) different from \( x \). Therefore, every point \( x \in (-1, 1) \) is an accumulation point of \( A \).
• Point \( 2 \)
  Consider the point \( 2 \). Every open neighborhood of \( 2 \) in \(\mathbb{R}\) is of the form \((2 - \epsilon, 2 + \epsilon)\) with \(\epsilon > 0\). The interval \((2 - \epsilon, 2 + \epsilon)\) with sufficiently small \(\epsilon\) does not contain any other points of \( A \) because \(\{2\}\) is an isolated point. Hence, \( 2 \) is not an accumulation point of \( A \).
• Endpoints of the Interval \((-1, 1)\)
  Consider the points \( -1 \) and \( 1 \). Every open neighborhood of \( -1 \) is of the form \((-1 - \epsilon, -1 + \epsilon)\) and contains points of \( A = (-1, 1) \), so \( -1 \) is an accumulation point of \( A \). Every open neighborhood of \( 1 \) is of the form \((1 - \epsilon, 1 + \epsilon)\) and contains other points of \( A = (-1, 1) \), so \( 1 \) is also an accumulation point of \( A \).
• Points Outside \([-1, 1] \cup \{2\}\)
  Consider a point \( x \notin (-1, 1) \cup \{2\} \).
  - If \( x < -1 \) or \( x > 2 \), there is a neighborhood of \( x \) that does not intersect \( A \), so \( x \) cannot be an accumulation point of \( A \).
  - If \( 1 < x < 2 \), any neighborhood of \( x \) will not contain \( 2 \) nor any point of \((-1, 1)\).

In conclusion, the accumulation points of \( A = (-1, 1) \cup \{2\} \) in the standard topology on \(\mathbb{R}\) are all the points in the closed interval \([-1, 1]\).

Therefore, the set of accumulation points of \( A \) is \([-1, 1]\).

And so on.