# Homeomorphic maps in the Euclidean space with the Euclidean metric

I am currently reading an introduction to topological and metric spaces and want to know whether the following statement is true:

Consider the Euclidean space $\mathbb{R}^n$ endowed with the Euclidean metric. Any function that maps an open ball in $\mathbb{R}^n$ to another open ball is homeomorphic.

It is clear to me that any function $f:X\rightarrow T$, with $X$ the discrete topology and T an arbitrary topology, is homeomorphic. Is the beforementioned statement somehow linked to this?

Thx

• Um: what do you mean by "homeomorphic"? The word "homeomorphic" is an adjective describing a relationship between two objects. So it doesn't make sense to say that "[A] function ... is homeomorphic." A function between two objects may be a "homeomorphism", if through it we can see that the two objects are homeomorphic. But given the third paragraph of your question, perhaps the adjective you are looking for is "continuous"? Sep 12, 2011 at 13:09
• It is not true that every map $f:X\rightarrow T$ with $X$ discrete is a homeomorphism; for one thing, the discrete topology is metrizable, so Hausdorff, and Hausdorff is a topological property, so if $T$ has any non-Hausdorff topology, then $f:X\rightarrow T$ cannot be a homeomorphism.
– gary
Sep 12, 2011 at 16:52
• People seem to be going to an awful lot of trouble to describe situations where $f:X\rightarrow T$ cannot be a homeomorphism when $X$ is discrete... isn't "T does not have the discrete topology" sufficient? Sep 12, 2011 at 18:15
• Well, I gave specific examples because I like to see specific examples in areas I'm not familiar with, and I imagined others in similar conditions would too.
– gary
Sep 12, 2011 at 18:22
• I am speculating here, but based on the OP's last paragraph I think he is confusing the terms "continuous function" and "homeomorphism". Sep 12, 2011 at 20:03

Consider the map $z \mapsto z^2$ in $\mathbb C = \mathbb R^2$. This maps the open unit ball continuously onto itself but is not a homeomorphism because it is not injective.
I think you may be referring to invariance of domain http://en.wikipedia.org/wiki/Invariance_of_domain , which says that if $U$ is ( U are?) an injective and continuous, then $U$ and $f(U)$ are homeomorphic.
And your second statement that any map $f: X\rightarrow T$ where $X$ is discrete, and $T$ has any topology is a homeomorphism , can fail in many ways: i)The discrete topology is totally-disconnected: take any p,q in your space X: then {p} and {X-p} is a disconnection. So if your $T$ is a connected topology, then f cannot be a homeomorphism. ii)Hausdorff: The discrete metric is metrizable, using the discrete metric $d(x,y)=1$ if $x \neq y$, and $0$ otherwise. iii)Metrizability: by metrizability of the discrete metric, if $T$ is not metrizable, $f$ cannot be a homeomorphism.