Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Suppose $f:X\to Y$ is a mapping from topological space $X$ to topological space $Y$.

Suppose $P$ is a partition of $X$. Do $f$ continuous over each element of $P$ and $f$ continuous over $X$ imply each other? Do we need $f$ to be continuous over the closure of each element of $P$, instead of just over each element of $P$?

Following is a sketch of the proof:

$f$ continuous over $X$ implies $f$ continuous over each element of $P$, because continuity over a set and continuity at each element of the set are equivalent.

For the same reason, For $f$ continuous over $X$ to imply $f$ continuous over (the closure of) each element of $P$, does it suffice to check if $f$ is continuous at each point at the boundaries of each element of $P$?

If one doesn't imply the other, what extra conditions are needed for the implication to hold?

Thanks and regards!

share|cite|improve this question
up vote 1 down vote accepted

Let $f:\mathbb{R}\to\mathbb{R}$ be continuous. Consider any partition of $\mathbb{R}$ and let $P$ be a cell of the partition. Then the function that equals $f+1$ on $P$ and $f$ on every other cell shows tha no partition is sufficient.

The space $\mathbb{R}^2$ partitioned into the lines $\mathbb{R}\times{x}$ for some $x\in\mathbb{R}$, shows that one can even partition a well-behaved space into perfect sets and get a counterexample.

Here is a rather weak positive result:

Let X be a metric space and consider a partition of $X$ such that each point has a neighborhood that meets only finitely many cells. Let $f$ be a function with values in a metric space that is continuous on the closure of each cell. Then $f$ is continuous.

Proof: Let $(x_n)$ be a sequence converging to a point $x$. Since a neighborhood of $x$ meets only finitely many cells, there is a cell that contains a subsequence of $(x_n)$ and its closure contains $x$. Since $f$ is continuous on the closure of this cell $f(x_n)\to f(x)$.

share|cite|improve this answer
+1. Thanks! In what sense does "weak" mean? – Tim Feb 8 '12 at 2:54
It requires a reather strong assumption. – Michael Greinecker Feb 8 '12 at 2:56
But I think in real analysis, the domains of a lot of piecewise functions I have seen though not many are partitioned in such a way that the strong assumption is met. – Tim Feb 8 '12 at 2:58
Thinking about it, the result holds for arbitrary topological spaces. The condition is really just making the pasting Lemma Benjamin Lim mentioned applicable locally. – Michael Greinecker Feb 8 '12 at 3:02
"the result holds for arbitrary topological spaces", you mean the pasting Lemma Benjamin Lim mentioned? – Tim Feb 8 '12 at 3:04

Requiring that $f$ be continuous on each element of the partition does not suffice - for example, consider the partition $$\mathbb{R}=(-\infty,0]\cup (0,\infty)$$ and the function $f:\mathbb{R}\to\mathbb{R}$ defined by $$f(x)=\begin{cases}0\text{ if }x\leq 0\\ 1\text{ if }x>0\end{cases}$$ which is continuous on each element of the partition, but not continuous overall.

Nor does requiring that $f$ be continuous on the closure of each element of the partition suffice - for example, $$\mathbb{R}=\bigcup_{a\in\mathbb{R}}P_a$$ where $P_a=\{a\}$ is a partition of $\mathbb{R}$, and for every $a\in\mathbb{R}$, $P_a=\overline{P_a}$. Any function $\mathbb{R}\to Y$ is continuous on each $P_a$, but there are many discontinuous functions with $\mathbb{R}$ as domain.

However, it is true that if $$X=\bigcup_{\alpha\in A}P_\alpha$$ is an open cover of $X$, i.e. each $P_\alpha$ is an open subset of $X$, then $f:X\to Y$ is continuous if and only if each restriction $f|_{P_\alpha}:P_\alpha\to Y$ is continuous. This is easy to see: suppose that $U\subseteq Y$ is an open subset. If each $f|_{P_\alpha}$ is continuous, then $$f|_{P_\alpha}^{-1}(U)=\{x\in P_\alpha\mid f(x)\in U\}=f^{-1}(U)\cap P_\alpha$$ is open in $P_\alpha$ (which has the subspace topology from $X$). Because each $P_\alpha$ is open in $X$, a subset of $P_\alpha$ is open in the subspace topology if and only if it is open as a subset of $X$. Thus, each $f|_{P_\alpha}^{-1}(U)$ is open as a subset of $X$, and hence $$f^{-1}(U)=f^{-1}(U)\cap X= f^{-1}(U)\cap\left(\bigcup_{\alpha\in A} P_\alpha \right)= \bigcup_{\alpha\in A}f|_{P_\alpha}^{-1}(U)$$ is an open subset of $X$. Thus, $f$ is continuous.

However, it is often the case that an open cover cannot also be a partition. A space is called connected when the only open cover that is also a partition is the trivial open cover, i.e. $\{X\}$. Intuitively, connected means exactly what you'd think - there aren't two "separate pieces". For example, $\mathbb{R}^n$, $\mathbb{S}^n$, and $\mathbb{T}^n$ are all connected, while $(0,1)\cup (2,3)$ is disconnected.

share|cite|improve this answer
+1. Thanks! What other conditions are needed? – Tim Feb 8 '12 at 1:36
(1) Thanks for mentioning the cover case. The reason I asked about partition instead of cover is because piecewise defined functions are not uncommon, and I think partition is more related to piecewise functions. So I am curious what conditions are sufficient and/or necessary for the partition case. (2) Note that in your first example, with same partition of $\mathbb{R}$, but requiring $f$ to be continuous over the closure of each element of the partition, is $f$ continuous over $\mathbb{R}$? – Tim Feb 8 '12 at 1:46
@Tim: That's true. Note that a topological space has a non-trivial open partition if and only if it is disconnected (that is essentially the definition of disconnected). You're also correct about point 2, but I can't think of any conditions on a partition that would make the statement "continuous on the closure of each element of a partition $\implies$ continuous" true in general. – Zev Chonoles Feb 8 '12 at 1:51
@Tim You may want to have a look at the pasting lemma. – user38268 Feb 8 '12 at 2:33
@BenjaminLim: Good to know! Thanks! – Tim Feb 8 '12 at 2:37

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.