How to show if a set is an open set in R

Question

I have been struggling to understand the concept of proving open sets. These Exercise Problems are in my lecture notebook, but they do not have solutions. I am trying to solve all the questions of Exercise 12.

It would be wonderful if someone can explain how to do each of the questions so I can have an idea when the test comes. I currently do not know how to approach the questions. I would be delighted if someone gives me simple tips and points to remember when solving these questions.

I am aware that question 12.4 is searchable on the internet, so it is not vital to answer. I just want to understand this concept. Thanks!

Here is the lecture notebook photo

Answer 1

As the definition has suggested, all you need to do to showing openness of a set $A$ is that for every $a\in A$, you have to look up an $\epsilon>0$ such that $(a-\epsilon,a+\epsilon)\subseteq A$, that $\epsilon>0$ may depend on $a$.

For the real line ${\bf{R}}$, this is open because for any $a\in{\bf{R}}$, just take $\epsilon=1$, and then $(a-1,a+1)\subseteq{\bf{R}}$ is trivial.

For the reason why $(0,1]$ is not open, intuitively, the problem arises at the point $1$, so assume that it were open, then $(1-\epsilon,1+\epsilon)\subseteq(0,1]$ by definition. But this is clearly not going through, because $1+\epsilon/2\in(1-\epsilon,1+\epsilon)$ but $1+\epsilon/2\notin(0,1]$. Note that $1+\epsilon/2>1$.

To show that $(0,1)\cup(2,3)$ is open. Given $a\in(0,1)\cup(2,3)$, then we have two cases.

Case I. $a\in(0,1)$, then take $\epsilon=\min\{1-a,a\}$, one claims that $(a-\epsilon,a+\epsilon)\subseteq(0,1)$: For $y\in(a-\epsilon,a+\epsilon)$, then $y-a<\epsilon<1-a$, so $y<1$, also that $y-a>-\epsilon>-a$, so $y>0$, hence $y\in(0,1)$.

Case II. $a\in(2,3)$, then take $\epsilon=\min\{3-a,a-2\}$, one claims that $(a-\epsilon,a+\epsilon)\subseteq(2,3)$ as well.

Answer 2

Some hints. These problems are about applying the definition, and correctly interpreting the set-theoretic notation.

1. You need to choose an arbitrary element of this set, say $x$, and come up with a recipe for finding an $\varepsilon$ such that $(x - \varepsilon, x + \varepsilon)$ is contained in this set. Note that since you have a union of two intervals, it might help to break the problem down into the case where $x \in (0, 1)$ and $x \in (2,3)$.

2. To prove that a set is not open, you need to find an example of an element where the condition fails.

3. This one might be easier if you re-write the set in new notation; you will see that it is very similar to question 1 then.

4. Once you have got this far, this one should be very easy.

Answer 3

Look at part 1. You need to prove that $S=(0,1)\cup(2,3)$ is open. Let $x$ be an element of $S$. Then either $x \in (0,1)$ or $x \in (2,3).$ Suppose $x\in (0,1).$ We need to show that for some $\varepsilon>0, (x-\varepsilon, x+\varepsilon)\subset S.$ Clearly it's enough to show $(x-\varepsilon, x+\varepsilon)\subset (0,1).$ This means $0 < x-\varepsilon < x+\varepsilon < 1.$ We just have to pick $\varepsilon$ small enough so that $\varepsilon < x < 1- \varepsilon.$

Similarly if $x \in (2,3).$

Answer 4

Let's say we have the set $A$ and $x\in A$, to show that $A$ is open we need to find $\epsilon$ such that $(x-\epsilon,x+\epsilon)\subseteq A$

How can we do it?

Let's look at the first example: $A=(0,1)\cup(2,3)$

Now if $x$ is in $(0,1)$ what $\epsilon$ needs to be? Notice that $x+(1-x)=1$ and $x-(x-0)=0$, so if $\epsilon$ is $\min((1-x),x)$ we are sure that $(x-\epsilon, x+\epsilon)\subseteq (0,1)\subset A$

Now we need to find epsilon for the case $x\in(2,3)$, we, we do the same: $x+(3-x)=3,\, x-(x-2)=2$ so if $\epsilon=\min((3-x),(x-2))$ we have $(x-\epsilon, x+\epsilon)\subseteq (2,3)\subset A$

Combine the two and we get:$\epsilon=\begin{cases}\min((1-x),x)&\text{if }x\in(0,1)\\\min((3-x),(x-2))&\text{if }x\in(2,3)\end{cases}$

So we have $\epsilon$ for every $x$ in $A$, hence, by definition, $A$ is open

Answer 5

I'm sure you've heard it before, but a set S in $\mathbb R$ is open if $\forall$ r $\in$ S $\exists$ a neighborhood around r (r-$\delta$,r+$\delta$) such that everything in the neighborhood is also in S. The neighborhood is just real numbers that are within some $\delta$ of r. So if your set is (1,2) then there is a neighborhood around 1.5 with $\delta$ = .1 where every real number within .1 of 1.5 is in (1,2). There is a $\delta$ that works for any number in (1,2) it turns out.
So how would you prove (1,2) is open? Well, take an arbitrary element r in (1,2). r is greater than 1 and less than 2. Take $\delta$ = min(2-r,r-1)/2. Let's think about what we know. We know that r + $\delta$ < 2 and r-$\delta$ > 1. We also know that all real numbers that are greater than 1 and less than 2 are in (1,2). Therefore $\forall$ r $\in$ (1,2) $\exists$ $\delta$ such that (r-$\delta$,r+$\delta$) is contained within (1,2). So we're done.

(r-$\delta$,r+$\delta$) is neighborhood around r.

Now let's prove (1,2] is not open. Take 2 $\in$ (1,2]. $\forall$ $\delta$ that are positive reals, (2-$\delta$,2+$\delta$) is not contained within (1,2], so the set is not open ($\delta$ is positive and real because the neighborhood has a positive size of 2$\delta$ and we're working with the reals.

The whole line is even easier than (1,2) in terms of openness; delta could be anything whereas for the (1,2) case, we had to choose a delta cleverly.

I'll leave the last one as an excercise.