It's possible that a set $X$ can be endowed with two or more topologies that are comparable. Over the years, mathematicians have used various  words to describe the comparison: a topology $\tau_1$ is said to be coarser than another topology $\tau_2$, and we write $\tau_1\subseteq\tau_2$, if every open set in $\tau_1$ is also an open set in $\tau_2$. In this scenario, we also say $\tau_2$ is finer than $\tau_1$. But other folks like to replace "coarser" by "smaller" and "finer" by "larger." Still others prefer to use "weaker" and "stronger." But how can we keep track of all of this? 

Here's another unspoken rule of mathematics: English doesn't always commute!

Word order is important...

In sum, the sufficient condition (a.k.a. the "if" direction) allows you to get what you want. That is, if you assume the sufficient condition, you'll obtain your desired conclusion. It's enough. It's sufficient. 

On the other hand, the necessary condition (a.k.a. the "only if" direction) is the one you must assume in order to get what you want. In other words, if you don't have the necessary condition then you can't reach your desired conclusion. It is necessary. 

Up to isomorphism” is a phrase that seems to get thrown around a lot without ever being explained. Simply put, we say two groups (or any other algebraic structures) are the same “up to isomorphism” if they’re isomorphic! In other words, they share the exact same structure and therefore they are essentially indistinguishable. Hence we consider them to be one and the same! But, you see, we mathematicians are very precise, and so we really don't like to use the word “same." Instead we prefer to say “same up to isomorphism.” Voila!

Here's a chart to help keep track of some of the different "flavors" of continuity in real analysis.

In short, "Noetherian-ness" is a property which generalizes "PID-ness." As Keith Conrad so nicely puts it, "The property of all ideals being singly generated is often not preserved under common ring-theoretic constructions (e.g. $\mathbb{Z}$ is a PID but $\mathbb{Z}[x]$ is not), but the property of all ideals being finitely generated does remain valid under many constructions of new rings from old rings. For example... every quadratic ring $\mathbb{Z}[\sqrt{d}]$ is Noetherian, even though many of these rings are not PIDs." (italics added)

In general, if $F$ is a field and $V$ is a vector space over $F$, the tensor product answers the question "How can I define scalar multiplication on $V$ by some larger field which contains $F$?" (Of course this holds if we replace the word "field" by "ring" and consider the same scenario with modules.)

Here's an algebra tip! Whenever you're asked to prove $$A/B\cong C$$ where $A,B,C$ are groups, rings, fields, modules, etc., mostly likely the The First Isomorphism Theorem involved!

If $X$ and $Y$ are normed vector spaces, a linear map $T:X\to Y$ is said to be <b>bounded</b> if $\|T\|< \infty$ where

$$\|T\|=\sup_{\underset{x\neq 0}{x\in X}}\left\{\frac{|T(x)|}{|x|}\right\}.$$

(Note that $|T(x)|$ is the norm in $Y$ whereas $|x|$ is the norm in $X$.) One can show that this is equivalent to

$$\|T\|=\sup_{x\in X}\{|T(x)|:|x|=1\}.$$ So intuitively (at least in two dimensions), we can think of $\|T\|$ this way…

Galois Theory is all about symmetry. So, perhaps not surprisingly, symmetries found among the roots of polynomials (via Galois theory) are closely related to symmetries of polygons in the plane (via geometry). In fact, the two are highly analogous! 

The Borel-Cantelli Lemma says that if $(X,\Sigma,\mu)$ is a measure space with $\mu(X)<\infty$ and if $\{E_n\}_{n=1}^\infty$ is a sequence of measurable sets such that $\sum_n\mu(E_n)<\infty$, then $$\mu\left(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k\right)=\mu\left(\limsup_{n\to\infty} En \right)=0.$$

(For the record, I didn't understand this when I first saw it (or for a long time afterwards). My only thought was, "But what does that mean? In English??") To help our intuition…

  In real analysis, there are two ways a measurable set $E$ can be small. Either

• the measure of $E$ is 0, OR
• $E$ is nowhere dense.

Intuitively, to say the measure of $E$ is $0$ means that...

Let a group $G$ act on a set $X$. The action is said to be transitive if for any two $x,y\in X$ there is a $g\in G$ such that $g\cdot x=y$. This is equivalent to saying there is an $x\in X$ such that $\text{orb}(x)=X$, i.e. there is exactly one orbit. And all this is just the fancy way of saying that $G$ shuffles all the elements of $X$ among themselves. In other words…

You're given a ring $R$ and are asked to show it's not a UFD. Where do you begin? One standard trick is to apply the Rational Roots Theorem….

An incomplete metric space is very much like a golf course: it has a lot of missing points!

Here's a measure theory trick: when asked to prove that a set of points in $\mathbb{R}$ (or some measure space $X$) has a certain property, try to show that the set of points which does NOT have that property has measure 0! This technique is used quite often.