Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I can't think of an example of a function from closed set of topological space to [-1,1] which cannot be extended to the whole of the space. Can anyone help me out with an example?

share|cite|improve this question

If $X$ is not normal, there are disjoint closed sets $A$ and $B$ that cannot be separated by disjoint open sets. Define $f$ from the closed set $A\cup B$ to $[-1,1]$ via $f(A)=1$ and $f(B)=-1$. Then as the preimage of a closed set is closed, $f$ is continuous. Suppose $g$ is a continuous extension of $f$. Then $g^{-1}[-1,0)$ and $g^{-1}(0,1]$ will be disjoint open sets containing $B$ and $A$ respectively, a contradiction.

It should be noted that the result of the Tietze Extension Theorem is a characterization of normal spaces.

share|cite|improve this answer

Here's an explicit example of David Mitra's answer in the simplest possible case.

Consider $X = \{a,b,c\}$ with $U\subseteq X$ open iff $U$ is empty or contains $a$. Then $\{b,c\}$ is a closed set, being a complement of $\{a\}$. Let $f(b) = 1$ and $f(c) = -1$. This is continuous because the subspace topology on $\{b,c\}$ is discrete.

This has no continuous extension to all of $X$. To see this, let $F:X\rightarrow [-1,1]$ be any set theoretic extension. By swapping $b$ and $c$, we may assume wlog that $F(a) \neq F(b)$. By Hausdorffness of $[-1,1]$, we may choose an open neighborhood $U$ of $[-1,1]$ containing $F(b)$ but not $F(a)$ or $F(c)$. Then $F^{-1}(U) =\{b\}$ which is not open, so $F$ isn't continuous.

share|cite|improve this answer

And here, to complement Jason's non-$T_1$ example, is a fairly simple completely regular example. Let $A$ be a set of cardinality $\omega_1$, $p$ a point not in $A$, and $X=A\cup\{p\}$. Topologize $X$ by making each point of $A$ isolated and giving $p$ a nbhd base of all sets of the form $X\setminus C$ such that $C$ is a countable subset of $A$. Let $Y=\Bbb N$, topologized by making each point of $\Bbb Z^+$ isolated and giving $0$ a nbhd base of sets of the form $\mathbb Z^+\setminus F$ such that $F$ is a finite subset of $\Bbb Z^+$. Finally, let $$Z=(X\times Y)\setminus\{\langle p,0\rangle\}$$ as a subspace of the product space $X\times Y$, let $H=A\times\{0\}$, and let $K=\{p\}\times\Bbb Z^+$.

Clearly $H$ and $K$ are disjoint closed subsets of $Z$. Suppose that $f:Z\to\Bbb R$ is continuous, and $f[K]=\{0\}$. Fix $n\in\Bbb Z^+$. For each $k\in\Bbb N$ let $$V_n(k)=f^{-1}\left[(-2^{-k},2^{-k})\right]\cap\Big(X\times\{n\}\Big)\;;$$ $V_n(k)$ is an open nbhd of $\langle p,n\rangle$ in $Z$, so there is a countable $C_n(k)\subseteq A$ such that $$V_n(k)=\Big(X\setminus C_n(k)\Big)\times\{n\}\;.$$ Now let $$\begin{align*}C&=\bigcup_{\langle n,k\rangle\in\Bbb Z^+\times\Bbb N}C_n(k)\text{ and}\\\\V&=X\setminus C\;;\end{align*}$$ $C$ is countable, so $V$ is an open nbhd of $p$ in $X$, and $f$ is constantly $0$ on $V\times\Bbb Z^+$. In particular, for any $x\in V\cap A$, $f(x,n)=0$ for all $n\in\Bbb Z^+$, so by continuity $f(x,0)=0$. Thus, $f$ must be $0$ on all but countably many points of $H$ and therefore cannot separate $H$ and $K$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.