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To speak of a property holding a.e. on a set $X$ you first need a measure. If $P(x)$ is a property defined for every point $x \in X$, then $P$ is true $\mu$-almost everywhere if $$\mu(\{ x \in X \mid \neg P(x)\}) = 0.$$ Perhaps a bit more precisely, you might require that there is a $\mu$-measurable set $N$ with the property that $\mu(N) = 0$ and $\{ x \in X ... 1 Here's an example to show that your concern about isolated points is justified. Let$f(x)=0$for all$x\in{\Bbb R}$and$g(x) = \max(x,0)$. These two functions are continuous and the are equal at (Lebesgue) a.e. point of$B:=(-\infty,0]\cup\{1\}$. But they are not identically equal on$B$. 1 It is valid only if the operator norm is induced by the Euclidean norm. If you use$||\cdot||_1$or$||\cdot||_{\infty}$, you'll get different expressions of the operator norm. In order to show that this equality holds when the norm on$\mathbb{R}^n$is the Euclidean norm, you can diagonalize$A^TA$in an orthonormal basis : there exists an orthogonal ... 1 (The question is basically answered in the comment) As a topological manifold is locally path connected, a connected manifold is automatically path connected, as a connected locally path connected topological space is path connected 1 Answering$1$and$2$simultaneously: It is equivalent due to the fact that$\overline{\mathbb{R}}$is first countable: that is, for every point$x$there exists a countable local basis. You can find more info on definitions here. Note that this is valid for an arbitrary metric space. However, the definition you state is not equivalent in general ... 1 While Aloizio Macedo correctly states that these two definitions are equivalent because of the first countability of the reals, here is a short proof: ($\Rightarrow$) Assume that$f$is continuous at$x\in T$with respect to the$\varepsilon-\delta$convention. Let$\{x_n\}$be a sequence in$T\setminus\{x\}$converging to$x$. Then fix$\varepsilon>0$. ... 1 You are right, there is nothing random in the definition. The point is that you can think of$\Omega$as "states of the world": you don't know what is going to happen in the future, what$\omega$is going to "materialise", so to speak. So you don't know which values$X$is going to take; but$X\$ gives you a way to "translate" events from the real world ...