## Return to Answer

Post Made Community Wiki by S. Carnahan♦
1

There is a very elegant proof that there exists no continuous injection from the plane into the real line. The proof can basically be given by drawing a picture on the blackboard.

Suppose there is such an injection $f$. Let $x$ and $y$ be distinct points in the plane and let $g_1$ and $g_2$ be paths from $x$ to $y$ such that $g_1(r_1)\neq g_2(r_2)$ for $r_1,r_2\in (0,1)$. Now this implies that $f\circ g_1((0,1))\cap f\circ g_2((0,1))=\emptyset$, contradicting the intermediate value theorem.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.934077262878418, "perplexity_flag": "head"}