Difference between the support of a discrete random variable and the atoms of its probability distribution

I'm confused about the difference between the support of a discrete random variable and the atoms of its probability distribution.

Suppose I have a discrete random variable $X$ defined on the probability space $(\Omega, \mathcal{F}, P)$ with support $\mathcal{A}\subseteq \mathbb{R}$. I am told that the distribution of $X$ is $H$ with $c$ atoms, $a_1,\ldots,a_c$.

Question: Is $\{a_1,\ldots,a_c\}=\mathcal{A}$?

My thought is the following: they are different because, while $\mathcal{A}$ could contain also a point $b$ at which $P(X=b)=0$, $P(X=a_i)>0$ for $i=1,\ldots,c$.

Is this correct?

A point $x$ is a member of the "support" of the probability distribution of $X$ precisely if for every open neighborhood $G$ of $x$, we have $\Pr(X\in G)>0$.
Now suppose $$X = \begin{cases} 1/2 & \text{with probability }1/2, \\[6pt] 1/3 \text{ or } 2/3 & \text{with equal probabilities totalling }1/4, \\[6pt] 1/4 \text{ or } 3/4 & \text{with equal probabilities totalling }1/8, \\ & (\text{We skipped 2/4 since it's not in lowest terms.}) \\[6pt] 1/5,\ 2/5,\ 3/5,\text{ or }4/5 & \text{with equal probabilities totalling }1/16, \\[6pt] 1/6\text{ or }5/6 & \text{with equal probabilities totalling }1/32, \\[6pt] 1/7,\ 2/7,\ 3/7,\ 4/7,\ 5/7,\text{ or }6/7 & \text{with equal probabilities totalling }1/64, \\[6pt] \text{and so on.} \end{cases}$$
Then the probability distribution of $X$ is discrete since the probabilities of the atoms add up to $1$, i.e. 100 percent of the probability is in point masses. The set of atoms is just the set of all rational numbers between $0$ and $1$.
But the support is the whole set $[0,1]$, which contains $0$ and $1$ (which are not atoms) and every irrational number between $0$ and $1$. The reason for that is for that every open interval about each such number, the probability that $X$ falls in that interval is positive.