Prove that the intervals $(a,\infty)$ and $(-\infty,b)$ are open

Question

Show that the intervals $(a,\infty)$ and $(-\infty,b)$ are open in $\mathbb{R}$.

I'm having trouble choosing a radius for every $x \in (a,\infty),\ y\in (-\infty,b)$ such that $B_1(x,r_1)\subset (a,\infty)$ and $B_2(y,r_2) \subset (-\infty,b)$ where both $B_1$ and $B_2$ are open balls.

Answer 1

For every $x\in (a,\infty)$, for any $y >0$ we have $x+y \in (a,\infty)$. Now, $a<x$, so, exists a $\delta >0$ such that $a<x+\delta\Rightarrow x+\delta \in (a,\infty)$. So, use that $\delta$, and you have the ball centered at $x$ with radius $\delta$ contained in $(a,\infty)$

Answer 2

Let's look at $(a, \infty)$. We want to show that for every $x \in (a,\infty)$ we can find an $r$ (based on $x$) so that $B(x,r) \subset (a, \infty)$.

Note two things:

1) $B(x,r) = (x-r, x+r)$

2) $(x-r, x+r)\subset (a,\infty) \iff a \le x-r$ and $x +r \le \infty$.

So we want $a \le x-r$ or $r \le x -a$.

And that's it. Any $r$ so that $0 < r \le x-a$ will do.

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Proof of 1 (if it isn't imediately obvious)

$B(x,r) = \{y\in \mathbb R: |x - y| < r\}$

$= \{y\in \mathbb R: -r < y - x < r\}$

$= \{y \in \mathbb R: x-r < y < x + r\}$

$= (x-r, x+r)$.

Restatement of 2: $(m,n) \subset (u,v) \iff u \le m$ and $v \ge n$.

Proof of 2 (if it's not immediately obvious)

$(m,n) =\{x \in \mathbb R| m < x < n\}$ so

$(m,n)\subset (u,v)$ means for all $x| m < x < n$ then $u < x < v$

Assuming neither $(m,n)$ nor $(u,v)$ are empty then there exists an $x_0$ so that $m < x_0 < y$ and $u < x_0 < v$. If $m < u$ then there is a $y$ so that $m < y < u < x_0 < n$ and $y \in (m,n)$ but $y \not \in (u,v)$ which contradicts $(m,n) \subset (u,v)$. So $m\ge n$. Similarly $n \le v$.

Answer 3

A variant:

• For $x\in (a,+\infty)$, consider the interval centred at $x$: $B(x,x-a)=(a,2x-a)$.

• For $x\in (-\infty,b)$, the interval $B(x, b-x)=(2b-x,b)$.

Note that both intervals have bounds $a$ (resp. $b$) and the symmetric of $a$ (resp. $b$) w.r.t. $x$.

Answer 4

Another way to see that $(a,\infty)$ and $(-\infty,b)$ are open, is to note that they can be written as a union of open sets, in fact $$(a,\infty)=\bigcup_{n\in\mathbb{N}}(a,a+n)$$ and $$(-\infty,b)=\bigcup_{n\in\mathbb{N}}(b-n,b) $$