# If a set is open, does it mean that every point is an interior point? [duplicate]

In Walter Rudin's Principles of Mathematical Analysis he defines open set as: "E is open if every point of E is an interior point of E." So this can be translated in logic as "If every point of E is an interior point of E, then E is open."

But does this mean that "If E is open, then every point of E is an interior point of E?" How is one sure when encountering these definitions that the converse also applies?

## marked as duplicate by Najib Idrissi, Community♦Jan 15 '18 at 16:15

• Yes. Definitions are ... definitions. So they are if and only if statements. If a GLIMP is be definition a thing that HIMPs then it isn't possible for something else to HIMP because by definition, that's what a GLIMP is. A thing that HIMPs. – fleablood Jan 15 '18 at 0:29
• Or to put is another way if we define if an open set doesn't have every point an interior point then... what does open mean? We can't say "If an animal bears it's young live then it is a mammal; that's a definition" and then say "but not all mammals bear their young live" because then ... what the heck was the definition of "mammal" if it wasn't that they bear young live??? – fleablood Jan 15 '18 at 0:37
• – ThePortakal Jan 15 '18 at 0:44
• @fleablood A set can contain all of its boundary points (which we call closed), none of them (open) or some of them. "Open" is just a word; it could have been used for this latter situation. – Kaz Jan 15 '18 at 4:36
• @Kaz Um.... and your point is? – fleablood Jan 15 '18 at 8:22

Definitions should always be treated as "if and only if". So when the text says something like "$E$ is open if every point of $E$ is an interior point of $E$" (I'm guessing it was italicized as such so as to indicate that the sentence is presenting a definition), read:

$E$ is open $\iff E$ every point of $E$ is an interior point of $E$.

Moreover, whenever you have an "if and only if" statement about an object, this statement can be used as a definition for that object. For instance, here are two possible (equivalent) definitions an author could choose for "infinite set" (there are surely many others):

$\bullet \quad$ We call a set $X$ infinite whenever there is an injection $\mathbb{N} \hookrightarrow X$.

$\bullet \quad$ We call a set $X$ infinite whenever there is a nonempty, proper subset $A \subsetneq X$ such that there is a bijection between $X$ and $X \setminus A$.

Such equivalent definitions are one tool that authors can use to motivate and ultimately present the same topic from different perspectives.

• Maybe, or maybe not, it should be pointed out that if you do have to equivalent definitions then one must prove the conditions are if and only if. In other words "Theorem: there is an injection for N to X if and only if whenever there is a proper subset A there is bijection between x and X\A Proof: blah blah blah" then "Definition: If X is a set where the conditions of the theorem hold then we call such a set "infinite"". Maybe. – fleablood Jan 15 '18 at 1:01
• For the sake of readers who haven't encountered them before, it might be worth noting that, while your example definitions of an infinite set are indeed equivalent in the nowadays fairly standard ZFC set theory, they are not necessarily equivalent in ZF set theory without the axiom of choice. – Ilmari Karonen Jan 15 '18 at 15:25

A definition by its nature is always an if and only if statement. We are identifying that "being open" and "having every point being an interior point" are the exact same thing.

A definition "A something is defined to be X if Y" isn't really a conditional logical statement in the something is true because it is a consequence-- it's more a statement that something is true because that's how it's labelled to be.

We can't state as a theorem "If an number is divisible by an even but not odd be power of 7 then it is a schlunknubt" if a "schlunknubt" isn't defined. And if that is our definition of "schlunknubt" then we can't then claim "but it might not be the case that all schlunknubts are divisible by an even but not odd be power of 7; we must prove it now" because ... if that isn't the case then what the heck is a schlunknubt after all? It hasn't actually been defined.