Why the Circle Cannot Embed in the Line

The Obstruction: Different Cut-Point Structures

An embedding preserves the topology of a space, including what happens when a point is removed. The circle and the real line have different behavior in this respect: removing any one point from the circle leaves a connected space, whereas removing an interior point from a nondegenerate interval disconnects it. The challenge is to show that the image of an embedded circle in the real line would have to be an interval with an interior point. Compactness supplies endpoints for the image, and connectedness rules out gaps between them.

We use the usual topology on \(\mathbb{R}\), and give \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) the subspace topology inherited from the Euclidean topology on \(\mathbb{R}^2\). A space is connected if it cannot be written as the union of two disjoint, nonempty open subsets. A cut point of a space \(X\) is a point \(x\in X\) for which \(X\setminus\{x\}\) is disconnected. An embedding \(f:X\to Y\) is a map that is a homeomorphism from \(X\) onto its image \(f(X)\), with the subspace topology on that image.

Why Deleting a Point Does Not Disconnect the Circle

First consider the point \(p_0=(1,0)\). Define a map \(h:\mathbb{R}\to S^1\setminus\{p_0\}\) by

The denominator \(t^2+1\) is positive for every \(t\in\mathbb{R}\). Also,

Thus \(h(t)\in S^1\). Its first coordinate is never \(1\), because \(t^2-1=t^2+1\) would imply \(-1=1\). Hence \(h(t)\ne p_0\).

Proof. The coordinate functions of \(h\) are continuous, since they are rational functions with denominator \(t^2+1\ne0\). For \((x,y)\in S^1\setminus\{p_0\}\), we have \(x\ne1\), so define \(g(x,y)=y/(1-x)\). This is continuous on \(S^1\setminus\{p_0\}\). From \(x^2+y^2=1\) and \(x\ne1\),

Writing \(t=g(x,y)\), we obtain \(t^2+1=2/(1-x)\). Therefore \((t^2-1)/(t^2+1)=x\) and \(2t/(t^2+1)=y\). So \(h(g(x,y))=(x,y)\). Conversely, substituting the coordinates of \(h(t)\) into \(g\) gives \(g(h(t))=t\). These continuous inverse maps prove that \(h\) is a homeomorphism. The real line is connected, so \(S^1\setminus\{p_0\}\) is connected.

For any \(p=(a,b)\in S^1\), the rotation \(R(x,y)=(ax-by,bx+ay)\) is a homeomorphism of \(S^1\) and sends \(p_0\) to \(p\). Its inverse is the rotation \(R^{-1}(x,y)=(ax+by,-bx+ay)\), since \(a^2+b^2=1\). Thus \(R\) carries \(S^1\setminus\{p_0\}\) onto \(S^1\setminus\{p\}\). A homeomorphism preserves connectedness, so \(S^1\setminus\{p\}\) is connected for every \(p\in S^1\). \(\square\)

Since \(S^1\) is path-connected, it is connected. Indeed, any two points can be written as \((\cos\theta,\sin\theta)\) and \((\cos\phi,\sin\phi)\) for real angles \(\theta,\phi\); the path \(s\mapsto(\cos((1-s)\theta+s\phi),\sin((1-s)\theta+s\phi))\), for \(s\in[0,1]\), joins them. We will use this connectedness as well as the stronger fact just proved about deleting a point.

Compact Connected Subsets of the Real Line

A subset \(K\) of a topological space is compact if every cover of \(K\) by open sets has a finite subcover. We use the Heine–Borel theorem from real analysis: a closed and bounded subset of \(\mathbb{R}^n\) is compact. We also need to know what compactness forces for a subset of \(\mathbb{R}\), so the relevant consequences are proved here.

Proof. The open sets \((-n,n)\), \(n\in\mathbb{N}\), cover \(K\), since every real number belongs to one of them. Compactness gives a finite subcover. If \(N\) is the largest index in that finite subcover, then \(K\subseteq(-N,N)\), so \(K\) is bounded.

To show that \(K\) is closed, fix \(x\in\mathbb{R}\setminus K\). For each \(y\in K\), let \(r_y=|x-y|/3>0\). The intervals \((y-r_y,y+r_y)\), as \(y\) ranges over \(K\), cover \(K\). Choose a finite subcover with centers \(y_1,\dots,y_m\). Let \(r=\min_{1\le i\le m}r_{y_i}>0\). The interval \((x-r,x+r)\) is disjoint from every chosen interval: if \(z\) belonged to both \((x-r,x+r)\) and \((y_i-r_{y_i},y_i+r_{y_i})\), then \(|x-y_i|\le |x-z|+|z-y_i|<r+r_{y_i}\le2r_{y_i}\), contradicting \(|x-y_i|=3r_{y_i}\). It follows that \((x-r,x+r)\cap K=\varnothing\). Thus every point outside \(K\) has an open neighborhood outside \(K\), so \(\mathbb{R}\setminus K\) is open and \(K\) is closed.

Because \(K\) is nonempty and bounded, its infimum \(m=\inf K\) and supremum \(M=\sup K\) exist in \(\mathbb{R}\). For every \(\varepsilon>0\), the definition of infimum gives \(k\in K\) with \(m\le k<m+\varepsilon\). Hence every open neighborhood of \(m\) meets \(K\). Since \(K\) is closed, this implies \(m\in K\): otherwise the open complement of \(K\) would contain a neighborhood of \(m\) disjoint from \(K\). The same argument, using the supremum, gives \(M\in K\). Therefore \(K\) has a minimum and a maximum. \(\square\)

Proof. By the preceding proposition, \(K\) has a minimum \(m\) and maximum \(M\), and \(m<M\) because \(K\) contains at least two points. Suppose some \(c\in(m,M)\) were not in \(K\). Then \(K\cap(-\infty,c)\) and \(K\cap(c,\infty)\) would be disjoint, nonempty, relatively open subsets of \(K\) whose union is \(K\). They are nonempty because \(m\in K\), \(M\in K\), \(m<c<M\), and \(c\notin K\). This contradicts connectedness. Thus every point between \(m\) and \(M\) belongs to \(K\); since \(K\subseteq[m,M]\), we have \(K=[m,M]\). Now fix \(c\in(m,M)\). The sets \([m,M]\cap(-\infty,c)\) and \([m,M]\cap(c,\infty)\) are disjoint, nonempty, relatively open in \(K\setminus\{c\}\), and cover that space. So \(K\setminus\{c\}\) is disconnected, and \(c\) is a cut point. \(\square\)

Proof That the Circle Cannot Embed in the Line

Proof. Suppose, for contradiction, that \(f:S^1\to\mathbb{R}\) is an embedding, and put \(K=f(S^1)\). The circle is a closed and bounded subset of \(\mathbb{R}^2\), so the Heine–Borel theorem gives that \(S^1\) is compact. A continuous image of a compact space is compact: if open sets cover \(f(S^1)\), their preimages under \(f\) form an open cover of \(S^1\); a finite subcover of \(S^1\) then gives a finite subcover of \(f(S^1)\). Hence \(K\) is compact.

The image \(K\) is also connected. If \(K\) had a separation into disjoint, nonempty, relatively open sets \(U\) and \(V\), then \(f^{-1}(U)\) and \(f^{-1}(V)\) would be disjoint, nonempty, open in \(S^1\), and would cover \(S^1\). This would contradict the connectedness of \(S^1\). Because \(f\) is injective and \(S^1\) has distinct points, \(K\) has at least two points. The proposition therefore gives \(K=[m,M]\) with \(m<M\). Let \(c=(m+M)/2\); then \(m<c<M\), so \(K\setminus\{c\}\) is disconnected.

Since \(c\in K=f(S^1)\), there is \(p\in S^1\) with \(f(p)=c\). The restriction of \(f\) maps \(S^1\setminus\{p\}\) homeomorphically onto \(K\setminus\{c\}\): it is the original homeomorphism onto \(K\), restricted to those points whose images are not \(c\). But \(S^1\setminus\{p\}\) is connected, while \(K\setminus\{c\}\) is disconnected. A homeomorphism preserves connectedness, giving a contradiction. Thus no such embedding exists. \(\square\)

Common Pitfall: An Injection Is Not Enough

The contradiction uses the embedding hypothesis, not merely the existence of a continuous injection into some space. An embedding must preserve the topology on the image, so it preserves connectedness after corresponding points are deleted. A continuous bijection need not be a homeomorphism in general, and a continuous injection need not identify the domain topology with the subspace topology on its image.

There is also a tempting but incorrect alternative argument: “The image is connected, so it is an interval, and therefore it has cut points.” Connectedness alone does not give endpoints. For example, \((0,1)\) is connected but has no minimum or maximum. In this proof it is compactness that supplies the endpoints; connectedness then fills every point between them. Both hypotheses are essential to the interval conclusion used here.

The obstruction can be summarized as a mismatch of cut points. Every point of the circle can be removed without disconnecting it; every compact connected subset of the line with more than one point has an interior point whose removal disconnects it. An embedding would preserve this feature, so the two spaces cannot be homeomorphic onto one another.