Question

true/false? prove.
1/2 is a limit of the interval ]− 2.5,1.5[

Answer 1

Answer on Question #69421 – Math – Topology

Question

true/false? prove.

1/2 is a limit of the interval ]- 2.5,1.5[

Solution

Recall that a limit point of a set $S$ in a topological space $X$ is a point $x$ that can be "approximated" by points of $S$ in the sense that every neighbourhood of $x$ with respect to the topology on $X$ also contains a point of $S$ other than $x$ itself.

Let $U$ be arbitrary neighbourhood of $\frac{1}{2}$. There is $\varepsilon > 0$ such that $\left(\frac{1}{2} - \varepsilon, \frac{1}{2} + \varepsilon\right) \subseteq U$. Put $a = \max \left\{-2.5, \frac{1}{2} - \varepsilon\right\}$. Note that, $a < \frac{1}{2}$. Consider a point $x$ such that $a < x < \frac{1}{2}$ (by example $x = \frac{a + \frac{1}{2}}{2}$). Since $a < x < \frac{1}{2}$, then $x \in \left(\frac{1}{2} - \varepsilon, \frac{1}{2} + \varepsilon\right) \subseteq U$ and $-2.5 < x < \frac{1}{2} < 1.5$, i.e. $x \in (-2.5, 1.5)$.

This means that $\frac{1}{2}$ is a limit of the interval $(-2.5, 1.5)$.

Answer: True.

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