The axioms of a topological space.

Question

The author of my text proves the properties of a topological space—(T1-3)—by

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But I don't really understand the statement in the proof of (T1) where he claims "...and, obviously, $X$ is open."

Is he saying that $X$ has to be open for that property to hold, or is he saying that he also proved that $X$ must be open. Because the former makes sense. Just making sure I'm thinking about it right.

Answer 1

Actually I think he means the latter -- in a metric space, a set $S$ is open, if for each its element, there is an open ball with positive radius centered at it, which is contained in $S$. Obviously, $X$ satisfies this definition, as every open ball is by definition a subset of $X$.

Answer 2

Notation. Let $(M,\rho)$ be a metric space. Fix $x\in M$. Fix real number $r >0$. $$B(x,r)$$ is notation for $$\{\xi\in M:\rho(x,\xi)<r\}.$$

Terminology. Let $(M,\rho)$ be a metric space. Fix $S\subseteq M$. Say $S$ is open if for each $s\in S$, there exists $\delta=\delta(s)>0$, such that $B(s,\delta)\subseteq S$.

Remark. Let $(M,\rho)$ be a metric space. Then $M$ is open. Indeed, for each $\xi\in M$, define $\delta(\xi)=1$ (any positive real works). Then $$B(\xi,\delta)\subseteq M.$$ As a result, $M$ is open. Notice: The ball only contains points in $M$. This is why $[0,1]$ is open when considering the metric space $([0,1],\rho)$, where $\rho(x,y)=|x-y|$.

Of course, $[0,1]$ is not open when considering the metric space $(\mathbb{R},\rho)$, where $\rho(x,y)=|x-y|$.

In short, being open is a relative terminology that depends on what space is being considered. This is why most mathematicians clarify that a set $S$ is open in $X$, where $S\subseteq X$ and $(X,\rho)$ is a metric space whose metric is understood via context; e.g. the metric induced by the established norm of a vector space.