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Let $\tau$ be the Euclidean topology defined on $\mathbb R$. If we define a set $S = (2,3) \cup (5,6)$. Then is the set $S$ an open set and open interval on $\tau$?

As per definition of open set,

A subset $A$ of $\tau$ is open set if $\forall x \in A$, $\hspace{5pt} \exists \hspace{3pt} a,b $ such that $x \in (a,b) \subseteq A$

As per this definition, we can find $(a,b) \in S$ such that $x \in (a,b)$. $x$ is any number in set $S$. So, set $S$ is open set.

I am not sure whether I correctly proved why set $S$ is open set. But I do not know how to prove set $S$ is open interval.

gyeox29ns
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3 Answers3

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$S$ is not an interval, open, or otherwise.

An interval $I$ in any ordered set has the property:

If $a<b<c$ and $a,c\in I$ then $b\in I$.

Your set $S$ is therefore open, but not an interval. It is the union of two open intervals.

Thomas Andrews
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The set $S$ is not an open interval.

An interval is a subset $A$ of $\Bbb R$ with the property that for all $x$ with $\inf\{a : a \in A\} < x < \sup\{a : a \in A\}$, we have that $x \in A$.

Since the infimum of $S$ is $2$ and the supremum is $6$, the number $4$, which is between these two, is not in $S$, so $S$ is not an interval.

By the way, every open interval is an open set. But as you saw in the example you provided, a disjoint union of open intervals is not itself an open interval (but it is an open set). Every open set can be expressed as an arbitrary union of open intervals, though.

layman
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Difference-1 if we cosider thee sequence with element of the subset of a open interval then the sequence is monotone increasing ......but if we consider a sequene with the element of the subset of an open set then the sequence is either monotone increasing or monotone decreasing .....

Gopal Ghosh
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