In Adams and Franzosa's Introduction to Topology, there are two ways that the standard topology on $\mathbb{R}^2$ is defined.
Definition 1. Let $d: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ be Euclidean distance, and for each $x \in \mathbb{R}^2$ and $\epsilon > 0$, let
$$B(x, \epsilon) = \{p \in \mathbb{R}^2: d(x, p) < \epsilon\}\text{.}$$
Let $$\mathcal{B} = \{B(x, \epsilon): x \in \mathbb{R}^2, \epsilon > 0\}\text{.}$$
Then $\mathcal{B}$ is a basis for the standard topology on $\mathbb{R}^2$.Theorem. Let $$\mathcal{B}^{\prime} = \{(a, b) \times (c, d) \subset \mathbb{R}^2: a < b, c < d\}\text{.}$$
Then $\mathcal{B}^{\prime}$ is a basis for the standard topology on $\mathbb{R}^2$.
In Figure 1.10 (p. 35), there is a drawing of a heart-shape object with dotted lines outlining it in $\mathbb{R}^2$. In Example 1.12 (p. 36), they aim to show such heart-shape objects are a basis for the standard topology on $\mathbb{R}^2$, and in showing this, they state
the hearts themselves are open sets in the standard topology.
Why is this the case? I suspect it's because hearts (as described above) can be written as a union of open rectangles, but I'm not certain.
If it helps, they say the same is also true for diamonds.
Best Answer
Thank you to Shubham Johri for pointing me in the right direction.
Theorem 1.4 of Adams and Franzosa states:
Under the basis $\mathcal{B}$ provided in the question, given any point in the heart, we may find a ball (which is open) which is centered at the point with radius $\epsilon > 0$ which is still contained in the heart, thus the hearts themselves are open in the standard topology.
It is important to note that the boundary outlining the heart is excluded from points that we look at, otherwise this would not hold.