# Does the function $f(x)=x^2$ admit a continuous extension to the Stone-Cech compactification?

Does the function $f(x)=x^2$ admit a continuous extension $\widetilde{f}:\beta\mathbb{R}\longrightarrow\mathbb{R}$ to the Stone-Cech compactification?

Proof. If $f$ admitted a continuous extension to the Stone-Cech compactification, then the triangle/maps would commute. The triangle being: $f = \widetilde{f} \circ g$ where $$f: \mathbb{R} \longrightarrow \mathbb{R} = \widetilde{f} \circ g: \mathbb{R}\longrightarrow \beta \mathbb{R}\longrightarrow\mathbb{R}$$ We have that $\widetilde{f}:\beta\mathbb{R}\longrightarrow\mathbb{R}$ is continuous. Also, $\beta\mathbb{R}$ is a compactification of $\mathbb{R}$ so $\beta\mathbb{R}$ is compact. Since the continuous image of compact sets is compact, then the image of $\widetilde{f}$ must be compact.

However, $f(x) = x^2$ is not bounded. Hence, not compact. So, the triangle doesn't commute. Hence, not every function $f$ admits a continuous extension $\widetilde{f}$ to the Stone-Cech compactification. End Proof.

Does this proof suffice?

• It looks like the actual question is something like "Does every function admit a continuous extension..", and you're proving that the answer to this question is no, by taking a counterexample $f(x) = x^2$. Is this correct? (The way you wrote it is very strange) – mathematician Aug 8 '16 at 17:55
• No the question is "Does every function $f(x) = x^2$ admit a continuous extension $\widetilde{f}:\beta\mathbb{R}\longrightarrow\mathbb{R}$ to the Stone-Cech compactification?" So, I said that because $f$ isn't bounded, then we won't get this conclusion. – Michael Aug 8 '16 at 18:00
• It was a question from an old PhD topology exam and it was worded this way. – Michael Aug 8 '16 at 18:03
• The question doesn't make sense though. Every function, but a single function is named? – mathematician Aug 8 '16 at 19:07
• For the question in your title, I would take it to mean an extension $\beta \mathbb R \to \beta\mathbb R$. But your formulation in the question itself is (as noted) easily seen to be false.. – GEdgar Aug 8 '16 at 20:09

A general argument: let $f : X \to \Bbb R$ be continuous and assume that $\tilde f : \beta X \to \Bbb R$ is a continuous extension of $f$. Since $\tilde f$ is continuous and $\beta X$ compact, it follows that $\tilde f$ must be bounded. Let $i : X \hookrightarrow \beta X$ be the natural injection. Notice that $\tilde f \circ i = f$. Since $\tilde f$ is bounded, it follows that $f$ too must be bounded. In words: every continuous function that admits an extension to the Stone–Čech compactification must be bounded.
Applying this to your exercise, since $(x \mapsto x^2) : \Bbb R^2 \to \Bbb R$ is not bounded, it may not be extended to $\beta \Bbb R$.