If every continuous function, let's say $f:X\to \mathbb{R}$, that they are uniformly continuous, can I assume that $X$ is compact?
I'm just wondering if I am true, can someone verify that I'm correct?
Since compact of a set is defined by it being closed and bounded, and since the domain $X$ has to be bounded and closed for the function $f$ to be uniformly continuous, I think it's obvious that the domain $X$ is compact.
The image $Y$ doesn't necessary have to be bounded, but the domain has to, right? Please point out if I'm wrong, or just say I'm right, yea!
Best Answer
First it is important to understand what uniform continuity means in a topological space $X$. In its full generality, one needs a uniform structure on $X$ to speak of uniform continuity, but let's stick to an easier concept ; a topological abelian group $X$ is an abelian group in which the operation $+$ is continuous as a function $+ : X \times X \to X$ (when $X \times X$ is equipped with the product topology).
Saying that $f : X \to \mathbb R$ is continuous in this context means that for every $\varepsilon > 0$ and for all $x \in X$, there exists $U_x \subseteq X$ open such that $x - y \in U_x$ implies $|f(x) - f(y)| < \varepsilon$.
Saying that $f : X \to \mathbb R$ is uniformly continuous in this context means that for every $\varepsilon > 0$, there exists $U \subseteq X$ open such that $x - y \in U$ implies $|f(x) - f(y)| < \varepsilon$, i.e. the open set $U$ does not depend on $x$ anymore.
If we take an arbitrary abelian group $X$ and equip it with the trivial topology (all its subsets are open), then of course this turns $X$ into a topological abelian group.
Take an arbitrary function $f : X \to \mathbb R$. Taking $U = \{0\}$ (where $0 \in X$ is the neutral element for addition), we see that $f$ is uniformly continuous, hence continuous. If the abelian group is infinite, the open cover that consists of a singleton for each point does not admit a finite subcover. More formally, $$ X = \bigcup_{x \in X} \{ x \} $$ is an open cover of $X$ which does not admit a finite subcover because $X$ is not finite. Therefore $X$ satisfies the property that every function $f : X \to \mathbb R$ is continuous, but $X$ is not compact.
Added : Even if $X \subseteq \mathbb R^n$ to put yourself in the context where compact is equivalent to closed and bounded, your guess is still wrong. Take $n = 1$ and see that as a subset of $\mathbb R$, $\mathbb Z$ is still endowed with the discrete topology, so that the proof above applies.
Hope that helps,