4

I have a slide that is just an itemize. I am using \vspace{.5cm} to space things out. For some reason between two of the bullets the spacing is larger than the others and I have to use \vspace{.015cm} to get a similar visual result.

How do I get even spaced bullets in a predictable way/what am I doing wrong?

My full slide is:

\documentclass{beamer}
\usetheme{Madrid}
\usepackage[utf8]{inputenc}
\usepackage{color}
\usepackage{mathtools}
\usepackage{multicol}
\usepackage{amsmath}
\usepackage[makeroom]{cancel}
\setlength{\columnsep}{1cm}
\DeclarePairedDelimiter\floor{\lfloor}{\rfloor}
\begin{frame}{Proof 2: Symmetry Preserving Maps}
\begin{itemize}
    \item  $\mathbb{Z}_2$ symmetry on $\mathbb{R}$: $x\longleftrightarrow -x$ 
    \vspace{.5cm}
    \pause
    \item $\mathbb{Z}_2$ symmetry on $S^d$: $x\longleftrightarrow -x$ 
    \vspace{.5cm}
    \pause
    \item $f_i:S^d\to \mathbb{R}$ given by $x\mapsto\mu_i(H^+(x))-\mu_i(H^-(x))$
    \vspace{.5cm}
    \pause
    \item $f_i(-x) = \mu(H^+(-x))-\mu(H^-(-x)) = \mu(H^-(x))-\mu(H^+(x)) = -f_i(x)$
    \vspace{.015cm}
    \pause
    \item $F:S^d\to\mathbb{R}^d$ given by $x\mapsto(f_1,f_2,\cdots,f_d)$
    \vspace{.5cm}
    \pause
    \item $F(-x) = -F(x)$
\end{itemize}
\end{frame}
\end{document}

using \vspace{0.015cm} between the offending pair: using <code>\vspace{0.015cm}</code> between the offending pair With \vspace{0.5cm} between each: With <code>\vspace{0.5cm}</code> between each Without any \vspace command between the offending pair: Without any <code>\vspace</code> command between the offending pair

  • Please, add a suitable document preamble in order to be able to reproduce your problem. – egreg Jul 30 '15 at 16:56
3
\documentclass[11pt]{beamer}
\usepackage[T1]{fontenc}
\usepackage{tgpagella}
\usetheme{Warsaw}

\begin{document}
    \begin{frame}{Proof 2: Symmetry Preserving Maps}
        \begin{itemize}[<+->]\setlength\itemsep{3ex}
            \item  $\mathbb{Z}_2$ symmetry on $\mathbb{R}$: $x\longleftrightarrow -x$ 
            \item $\mathbb{Z}_2$ symmetry on $S^d$: $x\longleftrightarrow -x$ 
            \item $f_i:S^d\to \mathbb{R}$ given by $x\mapsto\mu_i(H^+(x))-\mu_i(H^-(x))$
            \item $f_i(-x) = \mu(H^+(-x))-\mu(H^-(-x)) = \mu(H^-(x))-\mu(H^+(x)) = -f_i(x)$
            \item $F:S^d\to\mathbb{R}^d$ given by $x\mapsto(f_1,f_2,\cdots,f_d)$
            \item $F(-x) = -F(x)$
        \end{itemize}
    \end{frame}

\end{document}

enter image description here

  • This is a very simple solution. Thank you. – irh Jul 30 '15 at 17:13
2

It's a typical problem due to an unexpected spurious space.

The longer line ends almost at the boundary of the available space and there is a space (due to the end-of-line in the input) between the last $ and \vspace. Thus the paragraph consists of two lines, the second of which only contains \vspace.

When \vspace is issued in LR-mode (that is, when paragraphs are being formed) it inserts an invisible item in the paragraph. The best way to deal with \vspace is issuing it between paragraphs.

\documentclass{beamer}
\usetheme{Madrid}
\usepackage[utf8]{inputenc}
\usepackage{color}
\usepackage{mathtools}
\usepackage{multicol}
\usepackage{amsmath}
\usepackage[makeroom]{cancel}
\setlength{\columnsep}{1cm}
\DeclarePairedDelimiter\floor{\lfloor}{\rfloor}

\begin{document}

\begin{frame}{Proof 2: Symmetry Preserving Maps}
\begin{itemize}
    \item  $\mathbb{Z}_2$ symmetry on $\mathbb{R}$: $x\longleftrightarrow -x$ 

    \vspace{.5cm}
    \pause
    \item $\mathbb{Z}_2$ symmetry on $S^d$: $x\longleftrightarrow -x$ 

    \vspace{.5cm}
    \pause
    \item $f_i:S^d\to \mathbb{R}$ given by $x\mapsto\mu_i(H^+(x))-\mu_i(H^-(x))$

    \vspace{.5cm}
    \pause
    \item $f_i(-x) = \mu(H^+(-x))-\mu(H^-(-x)) = \mu(H^-(x))-\mu(H^+(x)) = -f_i(x)$

    \vspace{.5cm}
    \pause
    \item $F:S^d\to\mathbb{R}^d$ given by $x\mapsto(f_1,f_2,\cdots,f_d)$

    \vspace{.5cm}
    \pause
    \item $F(-x) = -F(x)$
\end{itemize}
\end{frame}
\end{document}

enter image description here

However, it's much better to use the available tools, instead of relying on manual spacing, like in Herbert's answer.

In my opinion beamer should also have a template for setting list related parameters.

\documentclass{beamer}
\usetheme{Madrid}
\usepackage[utf8]{inputenc}
\usepackage{color}
\usepackage{mathtools}
\usepackage{multicol}
\usepackage{amsmath}
\usepackage[makeroom]{cancel}
\usepackage{xpatch}
\setlength{\columnsep}{1cm}
\DeclarePairedDelimiter\floor{\lfloor}{\rfloor}

\xpatchcmd{\itemize}
  {\def\makelabel}
  {\usebeamertemplate{itemize body}\def\makelabel}
  {}{}

\defbeamertemplate*{itemize body}{default}{} % default is doing nothing

\setbeamertemplate{itemize body}{%
  \setlength{\itemsep}{0.5cm}%
}

\begin{document}

\begin{frame}
\frametitle{Proof 2: Symmetry Preserving Maps}

\begin{itemize}[<+->]
    \item  $\mathbb{Z}_2$ symmetry on $\mathbb{R}$: $x\longleftrightarrow -x$ 

    \item $\mathbb{Z}_2$ symmetry on $S^d$: $x\longleftrightarrow -x$ 

    \item $f_i:S^d\to \mathbb{R}$ given by $x\mapsto\mu_i(H^+(x))-\mu_i(H^-(x))$

    \item $f_i(-x) = \mu(H^+(-x))-\mu(H^-(-x)) = \mu(H^-(x))-\mu(H^+(x)) = -f_i(x)$

    \item $F:S^d\to\mathbb{R}^d$ given by $x\mapsto(f_1,f_2,\cdots,f_d)$

    \item $F(-x) = -F(x)$
\end{itemize}
\end{frame}

\end{document}

The output is the same.

1

enter image description here

\documentclass[11pt]{beamer}
\usetheme{Warsaw}
\usepackage{graphicx}
\usepackage{lmodern}
\usepackage{enumitem}
\setitemize{label=\usebeamerfont*{itemize item}%
\usebeamercolor[fg]{itemize item}
\usebeamertemplate{itemize item}}

\begin{document}
\begin{frame}{Proof 2: Symmetry Preserving Maps}
\begin{itemize}[itemsep=.5cm]
    \item  $\mathbb{Z}_2$ symmetry on $\mathbb{R}$: $x\longleftrightarrow -x$ 
    \pause
    \item $\mathbb{Z}_2$ symmetry on $S^d$: $x\longleftrightarrow -x$ 
    \pause
    \item $f_i:S^d\to \mathbb{R}$ given by $x\mapsto\mu_i(H^+(x))-\mu_i(H^-(x))$
    \pause
    \item $f_i(-x) = \mu(H^+(-x))-\mu(H^-(-x)) = \mu(H^-(x))-\mu(H^+(x)) = -f_i(x)$
    \pause
    \item $F:S^d\to\mathbb{R}^d$ given by $x\mapsto(f_1,f_2,\cdots,f_d)$
    \pause
    \item $F(-x) = -F(x)$
\end{itemize}
\end{frame}

\end{document}

Setting the vertical separation between items manually is not recommended besides being non-error free. enumitem package is typically used for this plus adjustment of many parameters of list environment. Bad news is that enumitem destroys the way beamer handles list environments and redefines labels and colors. So, we need to tell enumitem explicitly to keep beamer settings. This is done by the following piece of code:

\usepackage{enumitem}
\setitemize{label=\usebeamerfont*{itemize item}%
\usebeamercolor[fg]{itemize item}
\usebeamertemplate{itemize item}}

Finally, pass the option [itemsep=.5cm] to itemize:

  • Using enumitem with beamer is not at all recommended. – egreg Jul 30 '15 at 17:02
  • @egreg Yes, I gave that note "Bad news is that enumitem destroys ..." – AboAmmar Jul 30 '15 at 17:05
  • This is very helpful. I accepted Herbert's solution as it is what I ended up using, but the enumitem packege is very good to know about. Thank you! – irh Jul 30 '15 at 17:14

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