Connectedness of the Graph of $\frac{1}{x}$ and the $y$-axis

Question

Consider the topological space $G \cup Y,$ where $G = \{(x, \frac{1}{x}) \,:\, \mathbb{R} - \{0 \} \}$ and $Y = \{(0,y) \,:\, y \in \mathbb{R} \}$ with the subspace topology inherited from $\mathbb{R}^2.$ I am interested in whether or not this space is connected. Earlier, I convinced myself -- with the help of two colleagues -- that this space is indeed connected; however, our professor provided this proof that it is not connected.

Claim. $G \cup Y$ is not connected.

Proof. Consider the subsets $P = \{(x,y) \,:\, xy < \frac{1}{2} \}$ and $Q = \{(x,y) \,:\, xy > \frac{1}{2} \}$ of $\mathbb{R}^2.$ We note that both $P$ and $Q$ are open in $\mathbb{R}^2.$ Furthermore, we have that $G \subset Q$ since $xy = 1$ in $G$ and $Y \subset P$ since $xy = 0$ in $Y.$ We have therefore that $G \cup Y = (Q \cap G) \cup (P \cap Y)$ gives a nontrivial decomposition of the space $G \cup Y$ into clopen subsets, hence $G \cup Y$ is not connected.

But I cannot fathom why these intersections are clopen. From what I can tell, $Q \cap G$ is simply $G,$ and $P \cap Y$ is simply $Y.$ Furthermore, $G$ does not appear to be closed since it does not possess its limit points of $\pm \infty,$ and $Y$ does not appear to be open since its complement in $G \cup Y$ is not closed.

Could someone provide some insight into whether or not this space is connected?

Answer 1

(1) There are no limit points $\pm \infty$ here because there are no "points at infinity" in the set $\Bbb R^2$.

Limit points are defined within a space. You can always take a non-empty set $X$ and any topology $T_X$ on $X$ and take a point $p\not \in X,$ and define a topology on $Y=X\cup \{p\}$ such that the subspace topology on $X,$ as a subspace of $Y,$ is $T_X,$ and such that $p$ is a limit point of $X$ in the space $Y.$ But $p$ is a not a limit point of $X$ in the space $X$ because $p\not \in X.$

(2) The topology $T'$ on $G\cup Y,$ as a subspace of $\Bbb R^2,$ is $\{t\cap (G\cup Y):t\in T_{\Bbb R^2}\}$ where $T_{\Bbb R^2}$ is the set of all open subsets of $\Bbb R^2.$ This is the definition of "subspace topology".

$P$ and $Q$ are members of $T.$ For brevity let $P'=P\cap (G\cup Y)$ and $Q'=Q\cap (G\cup Y).$ So $P'$ and $Q'$ belong to $T'. $ That is, $P'$ and $Q'$ are open subsets of the $space $ $ G\cup Y.$

Furthermore $P'\cap Q'=\emptyset$ and $P'\cap Q'=G\cup Y.$ So the space $G\cup Y$ is the union of two non-empty disjoint open subsets ($P'$ and $Q'$), so it is not a connected space.

Answer 2

I'll share two ideas/sketches on why it isn't connected that I find more intuitive than your professor's.

1.

Consider the following subspace of $\mathbb{R}^2$:

$$\{(x-1,\frac{1}{x})\mid x<0\}\cup \{(x+1,\frac{1}{x})\mid x>0\}\cup (y\text{-axis})$$

(i.e. The graph of $\frac{1}{x}$ with it's negative part translated a unit to the left and it's positive part transaled a unit to the right, plues the $y$-axis.)

I imagine you'd agree this set is not connected. Now you only have to check that this set is homeomorphic to the unstralated graph and $y$-axis and you can safely and comfortably believe that this unstranslated graph + axis is not connected.

2.

Notice (at least visually) that for every point in the $y$-axis you can find a small open set that does not inersect the graph of $\frac{1}{x}$, and so the $y$-axis is an open set in the subspace. As it is also closed this is a non-trivial clopen set, i.e. the subspace is not connected.

Answer 3

G and Y are both closed sets within $R^2.$
Thus they are closed within the subspace S = G $\cup$ Y.
As inside S, G and Y are complements of each other,
they are open within S.
Whence S is the union of two disjoint, (relatively) open sets,
which assures S is disconnected.