A little background on me. I graduated with a degree in statistics, but I am returning to take a few more rigorous mathematics courses in hopes to pursue probability/statistics at a more rigorous level. I am currently having trouble getting comfortable proving that a set is open. My professor has given us some more complex (at least to me) examples, such as proving the set $$ \{f:[0,1] \longrightarrow \mathbb{R}; f \ is \ continuous\} $$ Along with other examples.

I do understand that in order to prove that a set is an open set, you need to prove/show that for each element in that set, there exists an open ball centred around that element that is contained in the set.

However, I would like to start proving sets are open with the most basic of examples (and I have been having trouble finding some online). Would anyone be able to provide some very basic examples of proving that a set is open, so I can gain some comfort with these types of proofs. For example, how would I prove that (a,b) is an open set by showing that for each $x \in (a,b)$, there exists $r>0$ such that $(x-r,x+r) \subset (a,b)$

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    $\begingroup$ Pick $r=\min\{x-a,b-x\}$ and contemplate why that works. $\endgroup$ Sep 23, 2020 at 19:43
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    $\begingroup$ In the last line need the containment symbol $\subset$ not $\in$. $\endgroup$
    – coffeemath
    Sep 23, 2020 at 19:48
  • $\begingroup$ For your set with functions, we need to know what the topology is. Presumably he's using $\|f\| = \max_{x\in [0,1]} |f(x)|$ to define a norm. $\endgroup$ Sep 23, 2020 at 21:10

1 Answer 1


Here's a detailed answer for you. Let $x \in (a,b)$ be arbitrary. Let $r= \frac{1}{2}\min\{x-a,b-x\}$. Let us show that the open ball $(x-r,x+r)$ is contained in $(a,b)$ Remember that $(x-r,x+r)$ is the set of all $y \in \mathbb{R}$ such that $x-r < y < x+r$. Thus, if $y \in (x-r,x+r)$ then $y-x < r <b-x$, so that $y < b$; analogously, $-r < y-x$. But $r< x-a$, so that $-r > -x+a$, so that $-x+a< -r < y-x$ and, consequently, $a < y$. We proved that $a< y < b$ and, thus, $y \in (a,b)$. Because $y$ was arbitrary, every $y \in (x-r,x+r)$ is also an element of $(a,b)$. In other words, $(x-r,x+r) \subset (a,b)$.

Note: the term 'open ball' is somewhat paradoxical in the present context because, well, you call $(x-r,x+r)$ open ball because you know that it is an open set, and this is because you know that open intervals are open sets. But this turns out to be a standard terminology in real analysis, even when the notion of open set is not yet defined.


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