# Proving that there is a path in set.

Given that $$\underline Y$$ is a subspace of $$\underline ℝ^2$$ and defined as follows:

$$I = [0,1]$$

$$X = (\{1\} × I) ∪ \left(I × \left(\{0\} \cup\left\{ \frac{1}{n} \,\Bigg\vert\, n \in \Bbb N \right\}\right)\right)$$

$$Y = X \setminus \{ (0.5, 0) \}$$

How can one prove that there is no path from $$(0,0)$$ to $$(1,0)$$ using sequences?

To show the contradiction I started by assuming that such a map exists:

$$f: \underline{I} \rightarrow \underline Y \ \text{continuous map}$$

$$f(0) = (0,0),\ \ f(1) = (1,0)$$

If $$f$$ is continuous, then it must hold that $$f^{-1}[\ [0,0.5)\times \{0\} ] = A$$ is a closed set. So $$A$$ contains its maximum $$m$$. From here on I'm not sure how to proceed. I wanted to somehow show that there is a sequence that due to $$m$$ being in $$A$$ cannot converge to $$f(m) = (x,0)$$.

You are on the right track.

So $$A$$ contains its maximum $$m\in[0,1)$$. On the other hand $$f(m)=(t,0)$$ for some $$0\leq t<0.5$$. Now take any sequence $$(x_n)\subseteq [0,1]$$ converging to $$m$$ from above, something like $$x_n=m+(1-m)/n$$. It follows that $$f(x_n)\to f(m)$$ and so if we write $$f(x_n)=(a_n,b_n)$$ then $$a_n<0.5$$ eventually, but since $$m$$ is the maximum of $$A$$ then $$b_n>0$$ (still convergent to $$0$$) whenever $$a_n<0.5$$.

It follows that the image of $$f$$ is not locally connected (around $$f(m)$$), by considering the open subset $$\{(x,y)\in im(f)\ |\ x<1\}$$, the projection onto $$Y$$-axis and preimages of clopen points. But any continuous function $$f:[0,1]\to X$$ into a Hausdorff space is a quotient map onto image. And being locally connected is preserved under quotients. This contradiction shows that no such continuous $$f$$ may exist.

Your space $$X$$ is a variant of the comb space. Many questions in this forum deal with it. Let us write $$X = \{1\} \times I \cup I \times N$$ with $$N = \{0\} \cup \left\{ \frac{1}{n} \,\Bigg\vert\, n \in \Bbb N \right\}$$. For later use we define $$N_m = \left\{ \frac{1}{n} \,\Bigg\vert\, n \ge m \right\}$$ and $$N^*_m = \left\{ \frac{1}{n} \,\Bigg\vert\, n < m \right\}$$.

My approach is this.

Consider a continuous $$f : I \to X$$. We claim that $$f(I)$$ intersects only finitely many of the sets $$B_n = [0,\frac{3}{4}] \times \{ \frac{1}{n}\}$$.
Otherwise there would exist a strictly increasing sequence $$(n_k)$$ of integers such that $$D_k = f(I) \cap B_{n_k} \ne \emptyset$$ for all $$k$$. Pick $$t_k \in I$$ such that $$f(t_k) \in D_k$$. The sequence $$(t_k)$$ has a convergent subsequence; so let us assume w.l.o.g. that $$t_k \to \tau \in I$$. By continuity $$f(t_k) \to f(\tau)$$. The second coordinate $$f_2(t_k)$$ of $$f(t_k)$$ is $$\frac{1}{n_k}$$, thus $$f_2(t_k) \to 0$$. Since the first coordinate $$f_1(t_k)$$ of $$f(t_k)$$ is always in $$[0,\frac{3}{4}]$$, we conclude that $$f(\tau) = (\theta,0)$$ for some $$\theta \in [0,\frac{3}{4}]$$. Obviuosly $$t_k \ne \tau$$ for all $$k$$. But $$U = [0,1) \times \left(\{0\} \cup\left\{ \frac{1}{n} \,\Bigg\vert\, n \in \Bbb N \right\}\right)$$ is an open neigborhood of $$f(\tau) = (\theta,0)$$ in $$X$$, thus there exists $$\epsilon > 0$$ such that $$f(t) \in U$$ for $$\lvert t - \tau \rvert < \epsilon$$. Choose $$k$$ such that $$\lvert t_k - \tau \rvert < \epsilon$$. Hence we get a path in $$U$$ connecting $$f(\tau)$$ and $$f(t_k)$$. This a contradiction because these points lie in different path components of $$U$$.

Now assume there is a path in $$Y$$ from $$(0,0)$$ to $$(1,0)$$. Then there also is a path $$f$$ from $$(0,0)$$ to $$(1,1)$$ because $$(1,0)$$ and $$(1,1)$$ lie in the same path component of $$Y$$. We know that $$f(I) \subset Y_m = \{1\} \times I \cup I \times N^*_m \cup [\frac{3}{4},1] \times N_m \cup (I \setminus \{\frac{1}{2} \}) \times \{0\}$$ for sufficiently large $$m$$. But $$Y' = \{1\} \times I \cup [\frac{3}{4},1] \times N \cup (I \setminus \{\frac{1}{2} \}) \times \{0\}$$ is a retract of $$Y$$, hence $$f$$ induces a path $$f'$$ in $$Y'$$ from $$(0,0)$$ to $$(1,1)$$. The coordinate function $$f_1'$$ is a continuos real-valued function such that $$f_1'(0) = 0, \ f_1'(1) = 1$$. The IVT says that $$f_1'(t) = \frac{1}{2}$$ for some $$t$$ between $$0$$ and $$1$$. This is a contradiction because $$Y'$$ does not contain any point whose first coordinate is $$\frac{1}{2}$$.