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

In the theorem 3.6 of Juhász's Cardinal Functions in General Topology appears the following symbol about sequence: $\frown$

The role context of it's appearance is the following:

Theorem. Let X be an compact topological space and $\kappa$ an cardinal such that $\chi(p,X)\geq \kappa$, for all $p\in X$. Then $|X|\geq 2^\kappa$.

Proof. First, set $\kappa =\omega$; we shall prove a little more than stated, namely that $X$ can be mapped continuously onto the interval $[0,1]$. To achieve this, we first define be induction on $n\in\omega$ an non-empty open subset $U_{\varepsilon}$ of $X$ for each finite sequence $\varepsilon\in 2^n$ in such way that

  1. $\overline{U_{\overset{\frown}{\varepsilon 0}}} \cup \overline{U_{\overset{\frown}{\varepsilon 1}}} \subseteq U_{\varepsilon}$
  2. $\overline{U_{\overset{\frown}{\varepsilon 0}}} \cap \overline{U_{\overset{\frown}{\varepsilon 1}}} = \emptyset$

The proof goes on...

What is the name of this symbol and what does it means?

share|cite|improve this question
up vote 2 down vote accepted

Usually this means extending the string $\varepsilon$ by appending $x$ to it at the end, or concatenating two strings.

So $(0,0,1)^\frown 0=(0,0,1,0)$.

share|cite|improve this answer

I suppose it means concatenation of sequence $\varepsilon$ with the one element $0$ or $1$.

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.