Elementary proof that knot complements are path-connected The complement of any (topological) knot is path-connected. More precisely, if $K$ is a subset of $\mathbb{R}^3$ (or $S^3$) homeomorphic to $S^1$, then $\mathbb{R}^3\setminus K$ (or $S^3\setminus K$) is path-connected.
This result is often assumed in introductory knot theory texts, and is an immediate consequence of Alexander Duality. I would like to cite this result in a first course in metric and topological spaces, however, so would like to know if a more elementary proof exists. By "elementary", I mean that I would like to avoid:


*

*Assuming that the knot is tame, and using any results from PL-topology;

*Assuming the knot is smooth, and using results from differential topology;

*Using results from algebraic topology (such as Alexander Duality).


Any hints or references would be appreciated.
 A: (citations for Włodzimierz's assertions)
Option 1: 
Cite Theorem IV (4) from Hurewicz-Wallman (p. 48): 
$E_n$ [i.e. $\mathbb{R}^n$] cannot be separated by a subset of dimension $\le n-2$.  The proof is based on Theorem IV (3) (see below) and some manipulation of rational 3-space in real 3-space, using only things one learns in a first course on general/metric topology. 
Option 2:


*

*Cite Theorem IV (3) (p. 44), which states that a necessary and sufficient condition that a subset $N$ of $E_n$, be $n$-dimensional is that $N$ contain a non-empty subset which is open in $E_n$. 

*Then an easy consequence is (Corollary 2) (p.46):  Let $U$ be an open set in $E_n$, which is neither empty nor dense,  and let $B$ be the boundary of $U$. Then $\text{dim } B = n – 1$. Apply this to $U$ being the complement of $K$ (assuming that $K$ separates $E_3$.

*Then argue that a continuous injection $Y$ of the circle $X$ into $\mathbb{R}^3$  cannot have dimension 2 by citing this Theorem proved by Hurewicz in 1933:  Suppose $f$ is a closed mapping of a (separable metrizable) space X on a (separable metrizable) space Y and suppose $\text{dim } Y - \text{dim } X = k,\text{ where } k > 0$. Then there is at least one point of $Y$ whose inverse-image contains at least $k + 1$ points.  In particular, dimension-raising maps on separable metric spaces aren't injections.

A: A proof may be given along the lines of the proof of the Jordan Curve theorem by Doyle (see this answer). This uses the fundamental group and a variation on Van Kampen, but not homology. So this probably still won't be satisfying to you, but it doesn't use Alexander duality (which is the normal way to do this). 
By removing a point from the knot, we may assume that we have a properly embedded line $L\subset \mathbb{R}^3$. Suppose that $\mathbb{R}^3-L$ is disconnected. Since $L$ is closed, $\mathbb{R}^3-L$ is open, and since it is disconnected, there are disjoint non-empty open subsets $U, V$ such that $U\cup V=\mathbb{R}^3-L$. Now embed $\mathbb{R}^3\subset \mathbb{R}^4$, with coordinates $(x,y,z,w)$, and $w=0$ gives $\mathbb{R}^3$. Then $\pi_1(\mathbb{R}^4-L)$ is non-trivial. This requires a generalization of Van Kampen's theorem which one can find, for example, in Peter May's book, Chapter 2. 
One may write $\mathbb{R}^4-L$ as a union of two simply connected open sets whose intersection is disconnected. These are $\{w>0\}\cup (\mathbb{R}^3-L) \times (-1,1)$ and $\{w<0\}\cup (\mathbb{R}^3-L) \times (-1,1)$. The intersection is $(\mathbb{R}^3-L) \times (-1,1)$, which is disconnected.  
Then the generalized Van Kampen's theorem implies $\pi_1(\mathbb{R}^4-L) \neq 0$. However, just as in Doyle's proof of the Jordan curve theorem, one may find a homeomorphism of $\mathbb{R}^4$ sending $L$ to the $w$-axis, whose complement is simply-connected (move L to its graph by a homeomorphism using the Tietze extension theorem, then project it to the $w$-axis; this has the same proof as Lemma 5.22 in Armstrong's book). This is a contradiction.
**Remark: ** One could use homology instead of fundamental group to carry out this argument, with Mayer-Vietoris replacing generalized Van Kampen. 

A: Not an answer to the question, but a hopefully related observation  to complete the proof and explain why the result is not obvious, which is may be of interest for your class. I  would mention that a point in $\mathbb{R}^3\setminus K$ may be such that every straight line for it meets $K$, that is, its sight is completely hidden by $K$ from outside the convex envelope of $K$. To make such example, just start from a  sphere-filling curve $\sigma:[0,1]\to\partial B(0,1)$, chosen so that for all $0<t<1$ one has $\sigma(t)\neq\sigma(0)=\sigma (1)$. Then define a simple closed curve $\kappa:[0,2]\to \mathbb{R}^3$ by $\kappa(t)=(2-t)\sigma(0)$ for $0\le t\le 1$ and $\kappa(t)=t\sigma(t)$ for $1\le t\le 2$; clearly the radial projection of this curve covers the whole unit sphere.
A: First construct a sequence Kn of tame embeddings of S1 in R3 that converges to K.
Then for any points A, B in S3-K construct a sequence of paths Pn from A to B such that Pn is in the compliment of Kn.  
Finally perform these constructions with Pn+1 pointwise so close to Pn that the sequence converges to an embedding P.
P is disjoint from every Kn, and hence disjoint from K, hence S3-K is path connected. 
