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Assume that $f : \mathbb{R} \to \mathbb{R}$ is continuous and $\lim_{x \to \pm \infty} f(x)$ exists. Can we argue that $f$ is uniformly continuous, by saying that a continuous extension of $f$ to the extended real-line (two-point compactification) is so?

EDIT: If the above is not possible, can we add some assumption on $f$ that makes it possible? For example, how about if we know that $f$ is monotone?

passerby51
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2 Answers2

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Short answer is 'no', slightly longer answer is 'no, but with some care the idea can be carried through.

The problem is that the two-point compactification is a topological notion while uniform continuity is a metric property (or more accurately, it is a property of uniformities (see uniform spaces)). In some more detail, to talk about uniform continuity of the two point compactification you need to consider an extension of the given metric to a metric on the compactification. There is more than one way to define such a metric that will agree with the topology so some care is needed.

Ittay Weiss
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  • Thanks. I understand that uniformity is not a topological property. (I have looked at uniform spaces before, but have not yet had enough motivation to read about them!) On the other hand, it seems that $|x-y|/(1+|x-y|)$ is equivalent to the usual metric of $\mathbb{R}$ (?) and can be extended to the compactification. – passerby51 Feb 07 '13 at 03:14
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A compact topological space $X$ admits a unique uniform structure $\mathfrak{U}$ compatible with its topology (for me, compact implies Hausdorff); moreover, $\mathfrak{U}$ consists of the neighborhoods of the diagonal $\Delta$ in $X\times X$ (for the product topology).

Moreover, any continuous function $f : X \to X'$ from $X$ as above to a uniform space $X'$ is uniformly continuous.

With the usual metric over $\mathbb{R}$, a fundamental system of entourages of its uniform structure is $V_{\epsilon} = \{(x,y) \in \mathbb{R}^2 \mid |x-y|< \epsilon \}$ with $\epsilon>0$. You can easily check that the uniform structure on $\mathbb{R}$ induced by $\overline{\mathbb{R}}$ is also the usual one.

Consequently, if $f : \mathbb{R} \to \mathbb{R}$ is continuous and extends to some $\overline{f} : \overline{\mathbb{R}} \to \mathbb{R}$ continuous, then $\overline{f}$ (and thus $f$) is uniformly continuous.

Notice that it is not true for any compactification of $\mathbb{R}$, in general the uniform structure induced by the compactification on $\mathbb{R}$ is strictly finer than the usual one. For a counterexample, build a continuous but non uniformly continuous function $f : \mathbb{R} \to [0,1]$ and extend it to $\tilde{f} : \beta \mathbb{R} \to [0,1]$.

However, it still works for Alexandroff compactification (aka one-point compactification).

Seirios
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  • Thanks. Are you using $\bar{R}$ for the two-point compactificaiton? I am still not quite comfortable with uniformities; so is the answer to my question yes? I.e., a function which is continuous on $\mathbb{R}$ and has limits at $\pm \infty$ automatically uniformly continuous on $\mathbb{R}$? Also, do you have a readable source for your first statement about compact top. spaces admitting a unique uniform structure? – passerby51 Feb 14 '13 at 15:03
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    @passerby51: Indeed, $\overline{\mathbb{R}}$ is the two-point compactification of $\mathbb{R}$. The answer to your question is yes, effectively, but there are more elementary proofs (without using $\overline{\mathbb{R}}$ and uniformities). As reference, I use General Topology by N. Bourbaki. – Seirios Feb 14 '13 at 17:53
  • I see. Thanks for your response. – passerby51 Feb 14 '13 at 19:26