Doubt in homeomorphism

Question

A mapping is a homeomorphism if it is a continuous, bijective map such that it's inverse is also continuous. This is the definition of homeomorphism which was taught to me. What does the following statement mean:"A point P is homeomorphic to an open disc in $\mathbb{R}^2$".

Answer 1

The statement says that there exists a continuous bijection from $\{p\}\to D\subseteq \Bbb{R}^2$ where $D$ is an open disk in $\Bbb{R}^2$ with continuous inverse. That is, there exists a homeomorphism $\phi:\{p\}\to D$. This is, of course, false. There is not even a bijection between $\{p\}$ and $D$. One of them has cardinality $1$, while the other has uncountable cardinality.

Answer 2

More generally, let $(X,\mathscr{T}_X)$ and $(Y,\mathscr{T}_Y)$ be topological spaces. We say that $(X,\mathscr{T}_X)$ and $(Y,\mathscr{T}_Y)$ are homeomorphic if there exists a homeomorphism $X\to Y$; that is, a bijection $\phi:X\to Y$ which is continuous (with respect to the topologies $\mathscr{T}_1$ and $\mathscr{T}_2$) and which has a continuous inverse $\phi^{-1}$. We often shorten this and say that $X$ and $Y$ are homeomorphic if the topologies are clear, and we often write $X\cong Y$ (or similar notations).

In your case, we are abbreviating the statement that $\big(\{x\},\mathscr{T}^\text{ind}_{\mathbb{R}^2}\big)\cong\big(\mathbb{R}^2,\mathscr{T}_{\mathbb{R}^2}\big)$ (already demonstrated by other answers to be false) as $\{x\}\cong\mathbb{R}^2$, where $\mathscr{T}_{\mathbb{R}^2}$ is the Euclidean topology on $\mathbb{R}^2$ and $\mathscr{T}^\text{ind}_{\mathbb{R}^2}$ is the induced topology on the point set $\{x\}\subseteq\mathbb{R}^2$.