Open Sets

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always room

Why do we call $(1, 5)$ an open interval on the real line or in the real numbers?

The reason is that each number in the interval feels like it is out in the open - that there is room around it. It may be a very small surrounding space around it, but room none-the-less.

We extend this idea that there is always room “around” each number to open sets in $\mathbb {R}$.

Open Sets in $\mathbb {R}$

In an open set in $\mathbb {R}$, each number has room on either side.

Open Set in $\mathbb {R}$

$S \subset \mathbb {R}$ is an open set in $\mathbb {R}$ provided

\[ \text {For each } a \in S, \text { there exists an } \epsilon > 0 \text { such that } (a - \epsilon , a + \epsilon ) \subset S \]

The idea is that inside an open set, every number sits inside an open interval inside the set. There is space between each number and the endpoint. There is a positive distance between each number and each endpoint.

Forcing a positive distance between each number and the endpoints turns out to have far reaching consequences, which we will investigate in this course. First, we need some language for open sets.

Open Intervals in $\mathbb {R}$

The open interval $(4,7)$ is an open set in $\mathbb {R}$.

Let $c$ be any number in $(4,7)$. Then the open subinterval $\left ( c - \frac {4+c}{2}, c + \frac {7+c}{2} \right )$ will always be inside $(4,7)$.

Therefore, for each $c \in (4, 7)$, there exists an open interval inside $(4,7)$, which contains $c$.

Since this is true for all of the numbers in $(4,7)$, the interval is an open set in $\mathbb {R}$.

The interval $(4,7]$ is not an open set in $\mathbb {R}$.

Consider the number $7$. Any open interval around $7$ would look like $(7 - \epsilon , 7 + \delta )$, with both $\epsilon > 0$ and $\delta > 0$. However, there is always a real number in the interval $(7, 7 + \delta )$, like $7 + \frac {\delta }{2}$, which lies outside $(4, 7]$.

Therefore, no matter which open interval you choose around $7$, it will always containing a number outside $(4, 7]$.

$(4, 7]$ is not an open set in $\mathbb {R}$.

$\blacktriangleright $ Open intervals cannot contain their endpoints.

Unions of Open Intervals

Any union of open intervals is itself an open set.

Let $S$ be a union of open intervals.
Let $c \in S$.
Then $c$ is in one of the open intervals.
Then there exists an $\epsilon > 0$, such that $(c - \epsilon , c + \epsilon )$ is a subset of this one open interval. This interval is a subset of $S$. Therefore, $(c - \epsilon , c + \epsilon ) \subset S$.

The Empty Set

The empty set, $\emptyset $, is the set with no element. The empty set is a subset of the real numbers. It is the subset containing no real numbers.

$\emptyset $ is an open set.

The reasoning is sort of reverse thinking. If $\emptyset $ is not open, then it MUST contain a number, such that there is no tiny open interval around it. There is no such number. Therefore, $\emptyset $ is an open set.

$\blacktriangleright $ These types of statements are often said to be vacuously true. They are true, because there is nothing that violates the definition - because there isn’t anything at all.

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more examples can be found by following this link
More Examples of Real-Valued Functions

2025-01-11 16:40:47