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I am working with the usual definition of a dense set, which is

Let $U$ be any non-empty open subset of $X$. A set $A$ is dense in $X$ iff $A \cap U \neq \emptyset$.

My highly informal and layman's interpretation of a dense set is this.

A set $A$ is dense in $X$ if for any $x\in X$, there is always some $a \in A$ that is nearby. For example, for any real number, there is always a 'nearby' rational number.

This intuition has not failed me so far, but I'd like to be sure that this will not cause me trouble/make things more difficult in other scenarios in the future.

Trogdor
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    This is a good intuition. I would replace "nearby" by "arbitrary close to $x$" though. – Dark Jun 20 '16 at 17:25
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    Your informal interpretation can be rigorously stated as "any neighborhood of $x$ intersects $A$". – Ningxin Jun 20 '16 at 17:25

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I think that one point is missing, that "nearby" is invariant under any definition of "nearness" (within the context of the given topology).

In the case of a metric space, this means that for any $\varepsilon$, we can find a rational number whose distance from $r$ is less than $\varepsilon$. So no matter what "near" means to you, because it's a fickle notion, you can still find rational numbers which are nearby.

To put this intuition to the test, think about a situation where you have a dense point. E.g., let $\tau$ be a topology on $\Bbb R$, and let $\tau_0=\{U\cup\{0\}\mid U\in\tau\}\cup\{\varnothing\}$. Then $0$ is a dense point in this topology. What does it mean? It means that for every neighborhood, of every point, $0$ is going to be there. It is omnipresent in every neighborhood. Why is this a test? Because it is very counterintuitive to our real numbers-based intuition about these notions of density.

Asaf Karagila
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Your intuition is correct. What is not correct is what you call the usual definition:

Let $U$ be any non-empty open subset of $X$. A set $A$ is dense in $X$ iff $A\cap U\neq \emptyset $.

Of course there is a problem of quantification. Instead

$A$ is dense in $X$ iff for any $U$ be any non-empty open subset, $A\cap U\neq \emptyset $

guestDiego
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