Notation for segments

 Great question — and your intuition is exactly right.

Is a part of the set \((a,b]\)?

No.
In the interval notation \((a,b]\):

• (a means the endpoint a is not included.
  
• b] means the endpoint b is included.
  

So the set \((a,b]\) contains all real numbers strictly greater than \(a\) and less than or equal to \(b\).

Formally:

\[
(a,b] = \{\, x \in \mathbb{R} : a < x \le b \,\}
\]

Since the condition is a < x, the point \(a\) does not satisfy it, so:

\[
a \notin (a,b].
\]

Intuition

You can think of it like a segment drawn with an open circle at \(a\) and a closed circle at \(b\):

a    -----● b

     ^

     no point here

Why this matters

This tiny difference (including or excluding a point) has big consequences for:

• Topology
  \((a,b]\) is not open, and not closed, but it’s still a valid 1‑D manifold with boundary (its boundary is the set \(\{a,b\}\), even though only \(b\) is included).
  
• Measure
  The length of \((a,b]\) is still \(b-a\). Removing a single point doesn’t change length.
  
• Analogies you were already making
  A non-inclusive endpoint behaves somewhat like a “missing point” — a kind of tiny defect — while the inclusive one doesn’t.
  

If you want, we can explore the topology of half‑open intervals further (they’re fascinating little objects).