Keeping track of some good counterexamples when needed. I’ve roughly filtered them by topic for my own convenience. Let me know if you spot any typos.

Real Analysis

Lebesgue null set that is also comeager

Let $Q_{k} = \bigcup_{q_n\in \mathbb{Q}}(q_n- 2^{-(n+k+1)}, q_n+2^{-(n+k+1)})$. Then, \(m(Q_k) \leq \sum_{n=1}^\infty 2^{-n-k} = 2^{-k}.\) Moreover, since $\mathbb{Q}\subseteq Q_k$, $Q_k$ is dense. It is also open since it is a union of open sets. Let $Q = \bigcap_{k\in\mathbb{N}}Q_k$, so that $m(Q) = 0$ and $Q$ is dense, by the Baire Category Theorem. Moreover, \(Q^\complement = \left( \bigcap_{k\in\mathbb{N}}Q_k \right)^\complement =\bigcup_{k\in\mathbb{N}}Q_k^\complement.\) Since $Q_k$ is open, $Q_k^\complement$ is closed. Also let $x\in Q_k^\complement$. Since $Q_k$ is dense, for all $\varepsilon > 0$ there exists $q\in Q_k$ such that $|x-q| < \varepsilon$, so $Q_k^\complement$ has empty interior. Thus, $Q^\complement$ is a meager subset of $\mathbb{R}$ whose complement has measure 0.

This is not really surprising once you realize that topological properties usually do not correspond well with measure-theoretic properties.

$\ell_0$, eventually zero sequences

This is a weird space. Firstly, $\ell_0\subseteq\bigcap_{p\in (0,\infty]}\ell^p$: it is $\ell^p$ for every single $p$. Also, every element of $\ell^p$ is a limit of a sequence in $\ell_0$ (for $p<\infty$). However, there is no norm on $\ell_0$ that makes it a complete vector space. This can be seen below.

Another weird property: $\ell_0$ has empty interior when viewed as a subset of $\ell^p$.

Also, when viewed as a subspace of $\ell^2$, there is no orthogonal projection onto $\ell_0$. To see this note that $\inf_{y\in\ell_0}\|x-y\|_2 = 0$ for all $x\in\ell^2$, but if $x\notin\ell_0$ then this infimum is not achieved. Alternatively note that if $x\in(\ell_0)^\perp$, then $x$ must be perpendicular to each of the canonical basis vectors, and thus $x = 0$. Yet another perspective on this is that $\mathrm{cl}(\ell_0) = \ell^2$, so $((\ell_0)^\perp)^\perp = \ell^2$.

This is an example of why closedness is necessary for orthogonal projections.

There is no slowest rate of decay of an absolutely convergent series

Suppose that there exists a sequence of positive numbers $\{a_n\}_{n\in\mathbb{N}}$ such that $\sum_{n\in\mathbb{N}}a_n|c_n| < \infty$ if and only if $\{c_n\}_{n\in\mathbb{N}}$ is bounded. Note that this implies that $A := \sum_{n\in\mathbb{N}}a_n <\infty$. Then, the map \(T:\{c_n\}_{n\in\mathbb{N}}\mapsto\{a_nc_n\}_{n\in\mathbb{N}}\) is an invertible linear map from $\ell^\infty$ to $\ell^1$. Also, \(\|Tx\|_1 = \sum_{n=1}^\infty a_n |x_n| \leq A\|x\|_\infty\) so $T$ is bounded. Therefore, $T^{-1}$ is bounded, since we are working with Banach spaces. Note that by the previous part $\ell_0$ is dense in $\ell^1$, and $T(\ell_0)= \ell_0$. However, $\ell_0$ is not dense in $\ell^\infty$, so such an isomorphism cannot exist.

More concretely, note that $\|\chi_{\{n\}}\|_1 = 1$, but \(\|T^{-1}(\chi_{\{n\}})\|_\infty = \|a_n^{-1}\chi_{\{n\}}\|_\infty = a_n^{-1}.\) Letting $n\to\infty$ we see that $T^{-1}$ is unbounded. This is a contradiction, so no such sequence $\{a_n\}_{n\in\mathbb{N}}$ can exist.

A vector space that cannot be Banach

Let $X$ be a a vector space of countably infinite dimension. That is, there exists $\{e_n\}_{n\in\mathbb{N}}$ such that for all $x\in X$ there exist unique $e_{n_1},\dots, e_{n_k}$ and $c_1,\dots,c_k$ so that $x = \sum_{j=1}^k c_j e_{n_j}$. Suppose there exists a norm on $X$ such that $X$ is complete. Then, consider the subspaces \(A_n := \mathrm{span}\{e_1,\dots,e_n\}.\) Note that $A_n$ is closed as it is finite dimensional. It also has empty interior, since $(A_n + \varepsilon e_{n+1}) \cap A_n = \varnothing$ for all $\varepsilon > 0$. Then, $X = \bigcup_{n\in\mathbb{N}}A_n$, but this contradicts the Baire Category Theorem.

Uniform Limit of Differentiable Function is not Differentiable

Define \begin{equation*} f_n(x)=\begin{cases} -1, & x \leq -\frac{1}{n} \\ nx, & x \in \left[-\frac{1}{n},\frac{1}{n}\right] \\ 1, & x \geq \frac{1}{n} \end{cases} \end{equation*}

Then, we define $F_n(x) = \int_{0}^x f_n(x)\,dx$, so that \begin{equation*} F_n(x) = \begin{cases} \frac{1}{2n}-x, & x\leq -\frac{1}{n} \\ \frac{n}{2}x^2, & x \in \left[-\frac{1}{n},\frac{1}{n}\right] \\ x - \frac{1}{2n}, & x\geq \frac{1}{n}. \end{cases} \end{equation*} Note that $F_n\in C^1$, and $|F_n(x) - |x| | \leq \frac{1}{n}$ for all $x$. Thus, $F_n\to |x|$ uniformly.

There are probably simpler examples.

$L^1\cap L^2_{\text{loc}} \not\subseteq L^2$

Consider \(f = \sum_{n=1}^{\infty} n\chi_{[n,n+\frac{1}{n^3}]}.\)

This is basically saying that being $L^1$ on the whole space and $L^2$ on every bounded set is not sufficient to be $L^2$

Linear Algebra

Diagonalizability and invertibility have no relation

Consider $\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$.

Matrix with no invariant subspace

The map from $\mathbb{R}^2\to\mathbb{R}^2$ defined by $\begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix}$ (rotation by $90^\circ$) has no nontrivial invariant subspace.

Algebraic completeness of the base field is important here.

Linear map whose image and kernel are equal

A silly example is just the zero map on the zero vector space. More generally, let $V$ be any vector space. Then, consider the map $T:V^2\to V^2$ defined by $(v_1,v_2)\mapsto(v_2,0)$. Its kernel is all vectors of the form $(v,0)$, which is also its image.

In $\mathbb{R}^2$ this looks like projection onto one axis and then reflection over the line $y=x$.

This isn’t a very interesting example but I remember this because it was a question on my first college math exam.

Group Theory

Nonabelian group where every subgroup is normal

Of course, an example of this is $Q_8$, the quaternion group. The nontrivial subgroups are $\{\pm 1\}, \{\pm 1, \pm i\}, \{\pm 1, \pm j\}$, and $\{\pm 1, \pm k\}$.

Almost as if this group was designed to be a counterexample.

Ring Theory

UFD that is not a PID

Consider $\mathbb{Z}[x,y]$ and the ideal $\langle x,y\rangle$. If this ideal was generated by a single element, then it would have to divide $x$, but then it must be $x$. However, $\langle x\rangle \lneq \langle x,y\rangle$.

Integral Domain that is not a UFD

This is $\mathbb{Z}[\sqrt{5}i]$, since $6 = (1+\sqrt{5}i)(1-\sqrt{5}i) = 2\cdot 3$.