[proofplan] We work in the affine chart $Z \neq 0$, where the curve is the real locus $y^2=f(x)$, and then account for the single point at infinity $O$. The sign pattern of the cubic determines the set of real $x$-coordinates over which real points exist. Over each such interval the real locus is obtained from the two continuous graphs $y=\pm\sqrt{f(x)}$, with the graphs meeting exactly at roots of $f$. In the one-root case those graphs form one connected affine branch whose projective closure contains $O$; in the three-root case they form one bounded oval and one unbounded branch, and the missing interval between them separates the two components. [/proofplan]

[step:Describe the affine real locus from the sign of the cubic]Let $A$ denote the affine real locus, namely $A=\{(x,y) \in \mathbb{R}^2 : y^2 = f(x)\}$. Let $\iota: \mathbb{R}^2 \to \mathbb{P}^2(\mathbb{R})$ be the affine chart map into the [real projective plane](/page/Projective%20Plane) given by $\iota(x,y)=[x:y:1]$. This map identifies $A$ homeomorphically with $E(\mathbb{R}) \setminus \{O\}$, because substituting $Z=1$ gives precisely $y^2=x^3+ax+b$, while $Z=0$ in the projective equation gives $X^3=0$, hence the only point at infinity is $[0:1:0]=O$. Define the derivative polynomial $f': \mathbb{R} \to \mathbb{R}$ by $f'(x)=3x^2+a$. Since $E(\mathbb{R})$ is nonsingular, the polynomial $f$ has no repeated real root: if $f(r)=0$ and $f'(r)=0$, then the affine defining function $F: \mathbb{R}^2 \to \mathbb{R}$, $F(x,y)=y^2-f(x)$, would satisfy $F(r,0)=0$ and have gradient $\nabla F(r,0)=(-f'(r),2\cdot 0)=(0,0)$, so $(r,0)$ would be a singular affine point of the equation $y^2=f(x)$. Thus, in the three-root case the roots are distinct. For any interval $I \subset \mathbb{R}$ on which $f(x)\geq 0$, define the upper graph map $\gamma_{I,+}: I \to \mathbb{R}^2$ by $\gamma_{I,+}(x)=(x,\sqrt{f(x)})$, and define the lower graph map $\gamma_{I,-}: I \to \mathbb{R}^2$ by $\gamma_{I,-}(x)=(x,-\sqrt{f(x)})$. Both maps are continuous because $f$ is continuous and the square-root map is continuous on $[0,\infty)$. Moreover, every point of $A$ lies on one of these graphs over a point $x$ with $f(x)\geq 0$.[/step]

[guided]The projective curve has one affine chart that contains all points except possibly points with $Z=0$. We therefore first isolate exactly what happens in that chart. Define \begin{align*} A = \{(x,y) \in \mathbb{R}^2 : y^2 = f(x)\}. \end{align*} Let $\iota: \mathbb{R}^2 \to \mathbb{P}^2(\mathbb{R})$ be the standard affine chart embedding into the [real projective plane](/page/Projective%20Plane) given by $\iota(x,y)=[x:y:1]$. On the cubic, substituting $Z=1$ into \begin{align*} Y^2Z = X^3 + aXZ^2 + bZ^3 \end{align*} gives exactly \begin{align*} y^2 = x^3 + ax + b = f(x). \end{align*} Thus $\iota$ identifies $A$ with the part of $E(\mathbb{R})$ lying in the chart $Z\neq 0$. Now we check the points with $Z=0$. If $[X:Y:0] \in E(\mathbb{R})$, then the projective equation becomes \begin{align*} 0 = X^3. \end{align*} Hence $X=0$, and the point is $[0:Y:0]$ with $Y\neq 0$. In real projective space all such points equal $[0:1:0]$, so the only point outside the affine chart is $O=[0:1:0]$. The real affine equation $y^2=f(x)$ has solutions over exactly those $x$ for which $f(x)\geq 0$. For such $x$, the two possible $y$-values are $\sqrt{f(x)}$ and $-\sqrt{f(x)}$. Therefore, on every interval $I$ where $f\geq 0$, the corresponding part of the curve is the union of the two continuous graphs of the maps $\gamma_{I,+}: I \to \mathbb{R}^2$, $\gamma_{I,+}(x)=(x,\sqrt{f(x)})$, and $\gamma_{I,-}: I \to \mathbb{R}^2$, $\gamma_{I,-}(x)=(x,-\sqrt{f(x)})$. This graph description is the main reduction: the topology of $E(\mathbb{R})$ is now controlled by the sign pattern of the cubic $f$ and by how these two graphs meet at roots of $f$.[/guided]

[step:Connect the projective curve in the one-root case] Assume $f$ has exactly one real root, and denote it by $r \in \mathbb{R}$. Since $f$ is a monic cubic, its sign is negative on $(-\infty,r)$ and positive on $(r,\infty)$; also $f(r)=0$. Hence \begin{align*} A = \gamma_{[r,\infty),+}([r,\infty)) \cup \gamma_{[r,\infty),-}([r,\infty)). \end{align*} The interval $[r,\infty)$ is [connected](/page/Connected%20Set), so both graph images are connected. They intersect at the point $(r,0)$ because $\sqrt{f(r)}=0$. Therefore $A$ is connected as the union of two connected subsets with nonempty intersection. It remains to account for $O$. Define two maps $\Gamma_+: [r,\infty) \to E(\mathbb{R})$ and $\Gamma_-: [r,\infty) \to E(\mathbb{R})$ by $\Gamma_+(x)=[x:\sqrt{f(x)}:1]$ and $\Gamma_-(x)=[x:-\sqrt{f(x)}:1]$. For $x>r$, divide the homogeneous coordinates by $\sqrt{f(x)}$ in the first map and by $-\sqrt{f(x)}$ in the second map. Since \begin{align*} \frac{f(x)}{x^3} \to 1 \quad \text{as } x\to\infty, \end{align*} we have \begin{align*} \frac{\sqrt{f(x)}}{x^{3/2}} \to 1 \quad \text{as } x\to\infty. \end{align*} Therefore \begin{align*} [x:\sqrt{f(x)}:1] &= \left[\frac{x}{\sqrt{f(x)}}:1:\frac{1}{\sqrt{f(x)}}\right] \longrightarrow [0:1:0]. \end{align*} Similarly, \begin{align*} [x:-\sqrt{f(x)}:1] &= \left[-\frac{x}{\sqrt{f(x)}}:1:-\frac{1}{\sqrt{f(x)}}\right] \longrightarrow [0:1:0]. \end{align*} Thus $O$ lies in the closure of $\iota(A)$ inside $E(\mathbb{R})$. Since $A$ is connected and $\iota$ is a homeomorphism onto its image, $\iota(A)$ is connected. The convergence above shows that $O \in \overline{\iota(A)}$, where the [closure](/page/Closure) is taken inside $E(\mathbb{R})$. Hence \begin{align*} \iota(A)\cup\{O\} \subset \overline{\iota(A)}. \end{align*} We now use the following elementary closure property of connected spaces. If $C$ is a connected subset of a [topological space](/page/Topological%20Space) $X$ and $C\subset B\subset \overline{C}$, then $B$ is connected. To prove this, suppose $B=U\cup V$ is a separation of $B$ by nonempty disjoint relative-open sets. Since $C$ is connected, $C$ lies in one of them, say $U$. Then $V$ is a nonempty relative-open subset of $B$ disjoint from $C$, so for some [open set](/page/Open%20Set) $W\subset X$ one has $V=B\cap W$ and $W\cap C=\varnothing$. But $V\subset B\subset\overline{C}$, contradicting that every open neighbourhood of a point of $\overline{C}$ meets $C$. Therefore every subset lying between a connected set and its closure is connected. Since \begin{align*} E(\mathbb{R})=\iota(A)\cup\{O\}, \end{align*} the space $E(\mathbb{R})$ is connected. [/step]

[step:Separate the bounded oval and the unbounded branch in the three-root case]Assume $f$ has three distinct real roots $r_1<r_2<r_3$. Since $f$ is monic and the roots are simple, its sign is negative on $(-\infty,r_1)$, positive on $(r_1,r_2)$, negative on $(r_2,r_3)$, and positive on $(r_3,\infty)$. Therefore the affine real locus is the disjoint union \begin{align*} A = A_{\mathrm{oval}} \cup A_{\infty}, \end{align*} where \begin{align*} A_{\mathrm{oval}} &= \gamma_{[r_1,r_2],+}([r_1,r_2]) \cup \gamma_{[r_1,r_2],-}([r_1,r_2]). \end{align*} Also define \begin{align*} A_{\infty} &= \gamma_{[r_3,\infty),+}([r_3,\infty)) \cup \gamma_{[r_3,\infty),-}([r_3,\infty)). \end{align*} The set $A_{\mathrm{oval}}$ is connected because it is the union of two connected graph images over $[r_1,r_2]$, and these two graph images meet at both endpoints $(r_1,0)$ and $(r_2,0)$. It is bounded because $x\in [r_1,r_2]$ and $|y|=\sqrt{f(x)}$ is bounded on the [compact](/page/Compact%20Space) interval $[r_1,r_2]$. In the topological sense, $A_{\mathrm{oval}}$ is a simple closed curve. Define $h:[0,2]\to A_{\mathrm{oval}}$ by \begin{align*} h(t)=(r_1+t(r_2-r_1),\sqrt{f(r_1+t(r_2-r_1))}) \quad \text{for } 0\leq t\leq 1, \end{align*} and \begin{align*} h(t)=(r_1+(2-t)(r_2-r_1),-\sqrt{f(r_1+(2-t)(r_2-r_1))}) \quad \text{for } 1\leq t\leq 2. \end{align*} Let $S^1$ denote the quotient of $[0,2]$ obtained by identifying $0$ and $2$. The two formulas agree at $t=1$, and $h(0)=h(2)=(r_1,0)$. Moreover $h$ is injective on $[0,2)$, since the upper and lower graphs intersect only where $f(x)=0$, namely at $x=r_1$ and $x=r_2$ on this interval. Hence $h$ descends to a continuous bijection $\tilde h:S^1\to A_{\mathrm{oval}}$. Since $S^1$ is compact and $\mathbb{R}^2$ is Hausdorff, $\tilde h$ is a homeomorphism onto its image. The set $A_{\infty}$ is connected because the two graph images over $[r_3,\infty)$ meet at $(r_3,0)$. As in the one-root case, both ends of $\iota(A_{\infty})$ converge to $O$ in $\mathbb{P}^2(\mathbb{R})$, so $O \in \overline{\iota(A_{\infty})}$, where the closure is taken inside $E(\mathbb{R})$. Since $\iota(A_{\infty})$ is connected and \begin{align*} \iota(A_{\infty})\subset \iota(A_{\infty})\cup\{O\}\subset \overline{\iota(A_{\infty})}, \end{align*} the closure property of connected sets proved above implies that \begin{align*} C_{\infty}:=\iota(A_{\infty})\cup\{O\} \end{align*} is connected. Also define \begin{align*} C_{\mathrm{oval}}:=\iota(A_{\mathrm{oval}}). \end{align*} Then $C_{\mathrm{oval}}$ is connected and bounded in the affine chart.[/step]

[guided]Now the cubic has three simple real roots $r_1<r_2<r_3$. Because the leading coefficient of $f$ is positive and each root is simple, the sign flips at each root. Thus $f$ is negative on $(-\infty,r_1)$, positive on $(r_1,r_2)$, negative on $(r_2,r_3)$, and positive on $(r_3,\infty)$. The equation $y^2=f(x)$ has real solutions exactly where $f(x)\geq 0$. Therefore real affine points occur precisely over the compact interval $[r_1,r_2]$ and the ray $[r_3,\infty)$. Define \begin{align*} A_{\mathrm{oval}} &= \gamma_{[r_1,r_2],+}([r_1,r_2]) \cup \gamma_{[r_1,r_2],-}([r_1,r_2]), \end{align*} and \begin{align*} A_{\infty} = \gamma_{[r_3,\infty),+}([r_3,\infty)) \cup \gamma_{[r_3,\infty),-}([r_3,\infty)). \end{align*} These sets are disjoint because their $x$-coordinates lie in disjoint subsets $[r_1,r_2]$ and $[r_3,\infty)$. They exhaust $A$ because there are no real points over the intervals where $f<0$. The set $A_{\mathrm{oval}}$ is connected: the upper and lower graphs over $[r_1,r_2]$ are connected images of the connected interval $[r_1,r_2]$, and they meet at both endpoints since \begin{align*} \sqrt{f(r_1)}=\sqrt{f(r_2)}=0. \end{align*} It is bounded because the $x$-coordinate lies in the bounded interval $[r_1,r_2]$, and the $y$-coordinate satisfies $|y|=\sqrt{f(x)}$, which is bounded on $[r_1,r_2]$ by continuity of $f$ on a compact interval. This bounded component is an oval in the topological sense of a simple closed curve. To see this, parametrize the upper graph from left to right and the lower graph from right to left. Define $h:[0,2]\to A_{\mathrm{oval}}$ by \begin{align*} h(t)=(r_1+t(r_2-r_1),\sqrt{f(r_1+t(r_2-r_1))}) \quad \text{for } 0\leq t\leq 1, \end{align*} and \begin{align*} h(t)=(r_1+(2-t)(r_2-r_1),-\sqrt{f(r_1+(2-t)(r_2-r_1))}) \quad \text{for } 1\leq t\leq 2. \end{align*} The two definitions agree at $t=1$ because both give $(r_2,0)$, and $h(0)=h(2)=(r_1,0)$. Let $S^1$ denote the quotient of $[0,2]$ obtained by identifying $0$ and $2$. The only intersections between the upper and lower graphs occur at the roots $r_1$ and $r_2$, so $h$ is injective on $[0,2)$. Thus $h$ factors through the quotient that identifies $0$ and $2$, giving a continuous bijection $\tilde h:S^1\to A_{\mathrm{oval}}$. Since $S^1$ is compact and $\mathbb{R}^2$ is Hausdorff, this continuous bijection is a homeomorphism onto its image. Therefore $A_{\mathrm{oval}}$, and hence $C_{\mathrm{oval}}$, is a bounded affine oval. The set $A_{\infty}$ is also connected: the two graphs over $[r_3,\infty)$ meet at $(r_3,0)$. This branch is unbounded in the affine chart, and its two ends approach the same projective point $O$. Indeed, for $x>r_3$, \begin{align*} \frac{f(x)}{x^3} \to 1 \quad \text{as } x\to\infty. \end{align*} Hence \begin{align*} [x:\sqrt{f(x)}:1] &= \left[\frac{x}{\sqrt{f(x)}}:1:\frac{1}{\sqrt{f(x)}}\right] \longrightarrow [0:1:0]. \end{align*} Similarly, \begin{align*} [x:-\sqrt{f(x)}:1] &= \left[-\frac{x}{\sqrt{f(x)}}:1:-\frac{1}{\sqrt{f(x)}}\right] \longrightarrow [0:1:0]. \end{align*} Hence $O \in \overline{\iota(A_{\infty})}$, where the closure is taken inside $E(\mathbb{R})$. The set $\iota(A_{\infty})$ is connected because $A_{\infty}$ is connected and $\iota$ is a homeomorphism onto its image. We now use the same closure property used in the one-root case: if $C$ is connected and $C\subset B\subset \overline{C}$, then $B$ is connected. Applying this with $C=\iota(A_{\infty})$ and $B=\iota(A_{\infty})\cup\{O\}$ shows that adjoining $O$ preserves connectedness. We therefore set \begin{align*} C_{\infty}:=\iota(A_{\infty})\cup\{O\}. \end{align*} Define also \begin{align*} C_{\mathrm{oval}}:=\iota(A_{\mathrm{oval}}). \end{align*} Both sets are connected.[/guided]

[step:Show that no additional connection appears in projective space]We prove that $C_{\mathrm{oval}}$ and $C_{\infty}$ are distinct [connected components](/page/Connected%20Component) of $E(\mathbb{R})$. Let $V_{\mathrm{oval}} \subset \mathbb{P}^2(\mathbb{R})$ be the affine-chart open set \begin{align*} V_{\mathrm{oval}}=\{[X:Y:Z]\in \mathbb{P}^2(\mathbb{R}) : Z\neq 0 \text{ and } X/Z<r_3\}. \end{align*} Let $q: \{[X:Y:Z]\in \mathbb{P}^2(\mathbb{R}) : Z\neq 0\}\to \mathbb{R}$ denote the affine coordinate function defined by \begin{align*} q([X:Y:Z])=\frac{X}{Z}. \end{align*} The set $V_{\mathrm{oval}}$ is open because it is the preimage of $(-\infty,r_3)$ under the continuous map $q$ on the open chart $Z\neq 0$. The sign pattern gives no real affine points with $r_2<x<r_3$, and the only projective point with $Z=0$ is $O$. Therefore \begin{align*} E(\mathbb{R})\cap V_{\mathrm{oval}}=C_{\mathrm{oval}}. \end{align*} Thus $C_{\mathrm{oval}}$ is open in the relative topology of $E(\mathbb{R})$. The set $A_{\mathrm{oval}}$ is compact because it is the union of two continuous images of the compact interval $[r_1,r_2]$. Hence $C_{\mathrm{oval}}=\iota(A_{\mathrm{oval}})$ is compact in $\mathbb{P}^2(\mathbb{R})$. Since $\mathbb{P}^2(\mathbb{R})$ is [Hausdorff](/page/Hausdorff%20Space), compact subsets are closed, so $C_{\mathrm{oval}}$ is closed in $E(\mathbb{R})$. Consequently \begin{align*} C_{\infty}=E(\mathbb{R})\setminus C_{\mathrm{oval}} \end{align*} is open in the relative topology of $E(\mathbb{R})$. The sets $C_{\mathrm{oval}}$ and $C_{\infty}$ are disjoint, nonempty, connected, and their union is all of $E(\mathbb{R})$: \begin{align*} E(\mathbb{R}) = C_{\mathrm{oval}} \cup C_{\infty}. \end{align*} Because they are complementary relative-open sets, no [connected set](/page/Connected%20Set) in $E(\mathbb{R})$ can meet both: such intersections would form a separation. Consequently $E(\mathbb{R})$ has exactly two connected components: the bounded affine oval $C_{\mathrm{oval}}$ and the component $C_{\infty}$ containing $O$.[/step]

[guided]We must still rule out a projective connection between the bounded oval and the unbounded branch. The point at infinity $O$ is attached only to the unbounded branch, but we need a topological separation argument to make this precise. First isolate the oval by an open affine condition. Define \begin{align*} V_{\mathrm{oval}}=\{[X:Y:Z]\in \mathbb{P}^2(\mathbb{R}) : Z\neq 0 \text{ and } X/Z<r_3\}. \end{align*} This set is open: the chart $Z\neq 0$ is open in $\mathbb{P}^2(\mathbb{R})$, and on that chart the coordinate function $q: \{[X:Y:Z]\in \mathbb{P}^2(\mathbb{R}) : Z\neq 0\}\to \mathbb{R}$ defined by \begin{align*} q([X:Y:Z])=\frac{X}{Z} \end{align*} is continuous, so $V_{\mathrm{oval}}$ is the preimage of the open interval $(-\infty,r_3)$. Intersecting with the curve gives exactly the oval. Indeed, the affine real points occur only over $[r_1,r_2]$ and $[r_3,\infty)$, there are no points over $(r_2,r_3)$, and the unique point with $Z=0$ is $O$, which is not in $V_{\mathrm{oval}}$. Therefore \begin{align*} E(\mathbb{R})\cap V_{\mathrm{oval}}=C_{\mathrm{oval}}. \end{align*} Thus $C_{\mathrm{oval}}$ is open in the relative topology of $E(\mathbb{R})$. Now we prove that the other piece is also relatively open without using the false claim that the set $\{Z=0\}\cup\{Z\neq 0, X/Z>r_2\}$ is open in projective space. The compactness of the oval gives the correct argument. The interval $[r_1,r_2]$ is compact, and the two graph maps $\gamma_{[r_1,r_2],+}$ and $\gamma_{[r_1,r_2],-}$ are continuous, so $A_{\mathrm{oval}}$ is compact as a finite union of compact images. The map $\iota$ is continuous, hence \begin{align*} C_{\mathrm{oval}}=\iota(A_{\mathrm{oval}}) \end{align*} is compact in $\mathbb{P}^2(\mathbb{R})$. Since $\mathbb{P}^2(\mathbb{R})$ is [Hausdorff](/page/Hausdorff%20Space), compact subsets are closed; therefore $C_{\mathrm{oval}}$ is closed in $E(\mathbb{R})$. Its complement \begin{align*} C_{\infty}=E(\mathbb{R})\setminus C_{\mathrm{oval}} \end{align*} is consequently open in the relative topology of $E(\mathbb{R})$. We have written $E(\mathbb{R})$ as the disjoint union of two nonempty connected relative-open sets: \begin{align*} E(\mathbb{R}) = C_{\mathrm{oval}} \cup C_{\infty}. \end{align*} A connected subset of $E(\mathbb{R})$ cannot meet both pieces, because its intersections with these two disjoint relative-open sets would form a separation. Hence these two connected sets are exactly the connected components. The first is the bounded affine oval, and the second contains $O$.[/guided]