In the Wikipedia's page of Tietze theorem is written that the first proof of this theorem was found for finite dimensional real vector spaces by Brouwer and Lebesgue, than for metric spaces by Tietze and finally for normal spaces by Urysohn.

My question is if there exist proofs of this theorem in the metric spaces setting that are easier than the general one (as for the Urysohn lemma) or even in the case of finite dimensional real vector space.


2 Answers 2


There is a "simpler" proof of the Tietze extension theorem for metric spaces, due to Hausdorff (1919): if $(X,d)$ is a metric space, $A\neq\varnothing$ is closed and $f:A\rightarrow[0,1]$ is continuous, then $\tilde{f}:X\rightarrow[0,1]$ given by $$\tilde{f}(p)=\begin{cases} f(p) & (p\in A) \\ \inf\Big\{f(q)+\frac{d(p,q)}{d(p,A)}-1\ \Big|\ q\in A\Big\} & (p\not\in A) \end{cases}\ ,\quad d(p,A)=\inf\{d(p,q)\mid q\in A\}$$ is continuous and clearly extends $f$. The quotes are due to the fact that the formula for the extension is more or less explicit (in fact, it reminds one of McShane's extension theorem for Lipschitz functions), but whether this proof can be considered simpler than Tietze's or not is up to discussion since it involves a fair amount of "epsilon-delta management".

It is easy to verify that $\tilde{f}$ takes values in $[0,1]$. Indeed, we have that $\frac{d(p,q)}{d(p,A)}-1\geq 0$ and $\tilde{f}(p)\leq\frac{d(p,q)}{d(p,A)}$ for all $p\in X\smallsetminus A$, $q\in A$, hence $\tilde{f}(p)\in[0,1]$ for all such $p$. The core of Hausdorff's proof is to establish the continuity of $\tilde{f}$. It follows from the definition of infimum of a nonvoid subset of $\mathbb{R}$ that $$\tilde{f}(p)\leq f(q)+\frac{d(p,q)}{d(p,A)}-1\ ,\quad p\in X\smallsetminus A\ ,\,q\in A$$ and $$\forall\delta>0\ ,\,p\in X\smallsetminus A\ \exists\ \!q\in A\text{ such that }f(q)+\frac{d(p,q)}{d(p,A)}-1<\tilde{f}(p)+\delta\ .$$ Particularly, the last assertion above entails that for such $q$ we have, since $\tilde{f}\leq 1$, that $$d(p,q)<(2+\delta)d(p,A)\ .$$ First we will establish continuity of $\tilde{f}$ in $A$. If $0<\delta<1$ and $p\in X\smallsetminus A$ is such that $d(p,A)<\frac{\delta}{4}$, we can find:

  • $q\in A$ such that $f(q)+\frac{d(p,q)}{d(p,A)}-1<\tilde{f}(p)+\delta$, hence $d(p,q)<3d(p,A)<\frac{3\delta}{4}$;
  • $q'\in A$ such that $d(p,q')<(1+\delta)d(p,A)<\frac{\delta}{2}$,

so that $$\tilde{f}(p)\leq f(q')+\frac{d(p,q')}{d(p,A)}-1<f(q')+\delta$$ and therefore $$f(q)-\delta<\tilde{f}(p)<f(q')+\delta\ .$$ Now, given $q_0\in A$, $\epsilon>0$, set $0<\delta<\min(\frac{\epsilon}{2},1)$ so that for all $r\in A$ with $d(r,q_0)<\delta$ we have $|f(r)-f(q_0)|<\frac{\epsilon}{2}<\epsilon-\delta$. If $d(p,q_0)<\frac{\delta}{4}$, we have that $q,q'$ as above satisfy $d(q,q_0)<\delta$ and $d(q',q_0)<\frac{3\delta}{4}$ by the triangle inequality, hence $|f(q)-f(q_0)|,|f(q')-f(q_0)|<\epsilon-\delta$ and therefore $$\begin{split}-\epsilon &<(\tilde{f}(p)-f(q))+(f(q)-f(q_0)) \\&=\tilde{f}(p)-f(q_0)\\&=(\tilde{f}(p)-f(q'))+(f(q')-f(q_0))\\ &<\epsilon\ .\end{split}$$ This establishes the continuity of $\tilde{f}$ in $q_0$. Since $q_0\in A$ was arbitrary, we conclude that $\tilde{f}$ is continuous in $A$. It remains to prove the continuity of $\tilde{f}$ in $X\smallsetminus A$ - let then $p_0,p\in X\smallsetminus A$, $0<\delta<\min(1,\frac{d(p_0,A)}{2})$ such that $d(p,p_0)<\delta$ and $q\in A$ such that $f(q)+\frac{d(p,q)}{d(p,A)}-1<\tilde{f}(p)+\delta$, hence $d(p,q)<3d(p,A)$. This entails that $$\begin{split} \tilde{f}(p_0) & \leq f(q)+\frac{d(p_0,q)}{d(p_0,A)}-1 \\ & <\tilde{f}(p)+\delta+\frac{d(p_0,q)}{d(p_0,A)}-\frac{d(p,q)}{d(p,A)} \\ &=\tilde{f}(p)+\delta+\frac{d(p_0,q)-d(p,q)}{d(p_0,A)}+d(p,q)\left(\frac{1}{d(p_0,A)}-\frac{1}{d(p,A)}\right) \\ &< \tilde{f}(p)+\delta\left(1+\frac{4}{d(p_0,A)}\right)\ .\end{split}$$ Notice that the inequality $\delta<\frac{d(p_0,A)}{2}$ was not used above, and the remaining hypotheses are symmetric in $p_0,p$. Hence, exchanging the roles of $p_0$ and $p$ (with a different $q$, adapted to $p_0$ instead of $p$), except on the choice of $\delta$, and then exploiting the fact that $$d(p,A)\geq d(p_0,A)-d(p,p_0)>d(p_0,A)-\delta>\frac{d(p_0,A)}{2}$$ by the triangle inequality, we obtain as well that $$ \tilde{f}(p)<\tilde{f}(p_0)+\delta\left(1+\frac{4}{d(p,A)}\right)<\tilde{f}(p_0)+\delta\left(1+\frac{8}{d(p_0,A)}\right)\ .$$ Both inequalities together establish the continuity of $\tilde{f}$ at $p_0$, as desired. Once more, $p_0\in X\smallsetminus A$ is arbitrary, hence $\tilde{f}$ is continuous in the whole of $X$.

The above proof can easily be modified to allow $f$ (and $\tilde{f}$) to assume values in any given compact interval of $\mathbb{R}$.

  • $\begingroup$ This is really an interesting answer, exactly of the type i was searching for. When I wrote "easier", more properly i meant "direct/using the metric structure". But there are thing in your proof that i don't think are correct, in particular from it appears the last "$=$" symbol onward. Can you check and if you think it's right expand/clarify a bit? $\endgroup$ Apr 12, 2021 at 22:48
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    $\begingroup$ Oops, there was a typo in the line just before the last equality above, just corrected. In the last equality above I used the fact that $$\frac{d(p_0,q)}{d(p_0,A)}-\frac{d(p,q)}{d(p,A)}=\frac{d(p_0,q)d(p,A)-d(p,q)d(p_0,A)}{d(p_0,A)d(p,A)}=\frac{d(p_0,q)d(p,A)-d(p,q)d(p,A)+d(p,q)d(p,A)-d(p,q)d(p_0,A)}{d(p_0,A)d(p,A)}=\frac{d(p_0,q)-d(p,q)}{d(p_0,A)}+d(p,q)\left(\frac{1}{d(p_0,A)}-\frac{1}{d(p,A)}\right)\ .$$ The last line is due to the fact that $d(p_0,q)-d(p,q)\leq d(p_0,p)<\delta$, $d(p,A)-d(p_0,A)\leq d(p,p_0)<\delta$ and $d(p,q)<3d(p,A)$. $\endgroup$ Apr 12, 2021 at 23:20
  • $\begingroup$ Now these passage are clear, but there is still a thing I don't get: why we can exchange the role of $p_0$ and $p$? It's seems to me that they have non-symmetrical role in this situation so we cannot exchange so easily. Can you explain this to me please? $\endgroup$ Apr 13, 2021 at 9:57
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    $\begingroup$ For the purpose of establishing this chain of inequalities, all we used was $p,p_0\in X\smallsetminus A$, $0<\delta<1$ and $d(p,p_0)<\delta$, which are clearly symmetric in $p_0,p$ The inequality $\delta<\frac{d(p_0,A)}{2}$ is not used until the very last inequality, after we exchanged the roles of $p$ and $p_0$, in which case the roles of $d(p_0,A)$ and $d(p,A)$ are also reversed. The chosen $q$ in the chain of inequalities is then also different. The last inequality follows from $d(p,A)\geq d(p_0,A)-d(p_0,p)>d(p_0,A)-\delta>\frac{d(p_0,A)}{2}$. I've added a clarifying remark to the answer. $\endgroup$ Apr 13, 2021 at 14:39
  • $\begingroup$ Thanh you for all your help, now all is clear and I accepted this answer. $\endgroup$ Apr 13, 2021 at 15:12

We actually did this in some form in an undergrad lecture on measure theory. To be precise we showed the following result (which was easier then the normal result^^):

Let $K$ be a compact metrical space, $C\subseteq K$ closed, and $f\colon C \to [-1,1]$. Then there exists $F\colon K \to [-1,1]$ such that $F|_C = f$.

We did this in multiple steps ($K$ being always compact, $A, B$ and $C$ being disjoint closed subsets):

1) Show that for all $A$, $B$ there is a continuous function $f\colon K \to [-1,1]$ and $g|_A = -1$, $g|_B = 1$. This can be done over the distance of a point from $A$ and $B$, using that both are closed and $K$ is compact for finiteness.

2) Let $f\colon C\to [-1,1]$ be continuous. Then there is a continuous $g\colon K\to [-2/3, 2/3]$ such that $|f-g| < 2/3 $ on $C$. For this set $A=f^{-1}([-1,-2/3])$ and $B=f^{-1}([2/3, 1])$ and apply 1) (and probably scale)

3) Show the result by building a sequence of functions $g_n$ ($g_0=f$) by repeatedly applying 2), always scaling with $2/3$ and taking the sum $\sum_{n=1}^\infty (2/3)^n g_n$.

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    $\begingroup$ So you're using the fact that Urysohn's Lemma is easy for metric spaces, and then using the usual deduction of Tietze's Extension Theorem from Urysohn's Lemma. I"m not convinced that you need compactness; Urysohn's Lemma is easy in any metric space; in particular, the distance from any point to any nonempty set is always finite. $\endgroup$ Nov 10, 2019 at 1:49
  • $\begingroup$ I had never seen the deduction of Tietze's Theorem from Urysohn in the general case, but I think that's exactly it. And yes, I also don't think we need compactness for all of this but it was just the form I had to hand. $\endgroup$
    – n314159
    Nov 10, 2019 at 14:06
  • $\begingroup$ If $\varnothing\neq A,B\subset X$, $A\cap B=\varnothing$ are closed subsets of a metric space $X$ with distance function $d$, an Urysohn function $g:X\rightarrow[0,1]$ is given by $$g(p)=\frac{d(p,A)}{d(p,A)+d(p,B)}\ .$$ Indeed, since $d(p,C)=\inf\{d(p,q)\mid q\in C\}$ is well defined and continuous for all $\varnothing\neq C\subset X$ and the denominator never vanishes since $A\cap B=\varnothing$, we get that $g$ is continuous and $0\leq f\leq 1$; moreover, $g|_A=0$ for $d(p,A)=0$ if $p\in A$ and $g|_B=1$ for $d(p,B)=0$ if $p\in B$. Set $f=2g-1$ to have values in $[-1,1]$ and you're done. $\endgroup$ Apr 10, 2021 at 7:44

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