Set vertical spacing between two particular items

There is a $(K_n)_{n\in\mathbb N}\subseteq E$ with
\begin{itemize}
\item $K_n$ is compact;
\item $K_n\subseteq\overset\circ K_{n+1}$; and
\item $d(\partial K_n,\partial K_{n+1})>0$
\end{itemize}
for all $n\in\mathbb N$.


This doesn't render well:

There's too much vertical empty space between the first and the second item (due to the circle above K). I would like to set the vertical spacing between these two items to 0pt, but I only know how I can set the vertical spacing between all items or between all items starting from a particular item.

So, how can we do this? Or would you recommend an other solution? Maybe it would be better to use only symbols which don't extend vertically inside a list. How is this usually done? Do you use $K^\circ$ instead? Or $\operatorname{int} K$? I don't like the latter and my problem with $K^\circ$ is that it's not consistent with other occurences in my document where I've used $\overset\circ K$.

• Adjust vertical space manually with \[-1.5ex] at the end of each \item. Change 1.5 to get desired results. – jak123 Jun 23 at 12:14
• @jak123 Why at the end of each item? – 0xbadf00d Jun 23 at 12:27
• Each item you would like to adjust. And its \\ not \ as I wrote before. – jak123 Jun 23 at 16:12

You can use \smash{\overset\circ K_{n+1}} to hide the height of that element.

\documentclass[a4paper]{article}
\usepackage{amsmath,amssymb}
\newtheorem{theorem}{Theorem}[section]
\begin{document}

\begin{theorem}
There is a $(K_n)_{n\in\mathbb N}\subseteq E$ with
\begin{itemize}
\item $K_n$ is compact;
\item $K_n\subseteq\smash{\overset\circ K_{n+1}}$; and
\item $d(\partial K_n,\partial K_{n+1})>0$
\end{itemize}
for all $n\in\mathbb N$.
\end{theorem}

\end{document}


• Perfect, thank you very much. – 0xbadf00d Jun 23 at 12:57

You don't have this problem if you use the \mathring accent:

\documentclass[a4paper]{article}
\usepackage{amsmath,amssymb}
\newtheorem{theorem}{Theorem}[section]

\begin{document}

\begin{theorem}
There is a $(K_n)_{n\in\mathbb N}\subseteq E$ with
\begin{itemize}
\item $K_n$ is compact;
\item $K_n\subseteq \ring{K}_{n+1}$; and
\item $d(\partial K_n,\partial K_{n+1})>0$
\end{itemize}
for all $n\in\mathbb N$.
\end{theorem}

\end{document}